primitive recursion - department of mathematicsorion.math.iastate.edu/jdhsmith/math/jgknf4cf.pdf ·...
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Primitive recursion
DefinitionGiven a binary function β(m, n) and any a ∈ N define the functionϕ(n) as follows;
ϕ(0) := a and ϕ(n + 1) := β(n, ϕ(n)).
I Note how the successor function σ0(n) = n + 1 is builtdirectly into the definition of primitive recursion.
I The class PR is built up from σ0 the constant functions andthe projections πi (a1, ..., ai , ..., an) = ai by substitution andprimitive recursion.
DMACC John M. Gillespie 1
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Primitive recursion
DefinitionGiven a binary function β(m, n) and any a ∈ N define the functionϕ(n) as follows;
ϕ(0) := a and ϕ(n + 1) := β(n, ϕ(n)).
I Note how the successor function σ0(n) = n + 1 is builtdirectly into the definition of primitive recursion.
I The class PR is built up from σ0 the constant functions andthe projections πi (a1, ..., ai , ..., an) = ai by substitution andprimitive recursion.
DMACC John M. Gillespie 2
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Primitive recursion
DefinitionGiven a binary function β(m, n) and any a ∈ N define the functionϕ(n) as follows;
ϕ(0) := a and ϕ(n + 1) := β(n, ϕ(n)).
I Note how the successor function σ0(n) = n + 1 is builtdirectly into the definition of primitive recursion.
I The class PR is built up from σ0 the constant functions andthe projections πi (a1, ..., ai , ..., an) = ai by
substitution andprimitive recursion.
DMACC John M. Gillespie 3
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Primitive recursion
DefinitionGiven a binary function β(m, n) and any a ∈ N define the functionϕ(n) as follows;
ϕ(0) := a and ϕ(n + 1) := β(n, ϕ(n)).
I Note how the successor function σ0(n) = n + 1 is builtdirectly into the definition of primitive recursion.
I The class PR is built up from σ0 the constant functions andthe projections πi (a1, ..., ai , ..., an) = ai by substitution
andprimitive recursion.
DMACC John M. Gillespie 4
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Primitive recursion
DefinitionGiven a binary function β(m, n) and any a ∈ N define the functionϕ(n) as follows;
ϕ(0) := a and ϕ(n + 1) := β(n, ϕ(n)).
I Note how the successor function σ0(n) = n + 1 is builtdirectly into the definition of primitive recursion.
I The class PR is built up from σ0 the constant functions andthe projections πi (a1, ..., ai , ..., an) = ai by substitution andprimitive recursion.
DMACC John M. Gillespie 5
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Two surprisingly useful functions
We define by primitive recursion sg and s̄g.
I sg(0) = 0 with sg(n + 1) = 1.
I s̄g(0) = 1 with s̄g(n + 1) = 0.
I Exponent notation is used e.g. 11sg = 1 and 2s̄g = 0.
DMACC John M. Gillespie 6
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Two surprisingly useful functions
We define by primitive recursion sg and s̄g.
I sg(0) = 0 with sg(n + 1) = 1.
I s̄g(0) = 1 with s̄g(n + 1) = 0.
I Exponent notation is used e.g. 11sg = 1 and 2s̄g = 0.
DMACC John M. Gillespie 7
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Examples
Well known primitive recursive functions. Takes some work!
1. The n-th prime pn
2. The exponent expl(n) of the l-th prime in the primefactorization of n ∈ N.
3. The residue res(a,n) of a modulo n.
4. Restricted subtraction a ·− b.
DMACC John M. Gillespie 8
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Examples
Well known primitive recursive functions. Takes some work!
1. The n-th prime pn
2. The exponent expl(n) of the l-th prime in the primefactorization of n ∈ N.
3. The residue res(a,n) of a modulo n.
4. Restricted subtraction a ·− b.
DMACC John M. Gillespie 9
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Examples
Well known primitive recursive functions. Takes some work!
1. The n-th prime pn
2. The exponent expl(n) of the l-th prime in the primefactorization of n ∈ N.
3. The residue res(a,n) of a modulo n.
4. Restricted subtraction a ·− b.
DMACC John M. Gillespie 10
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Examples
Well known primitive recursive functions. Takes some work!
1. The n-th prime pn
2. The exponent expl(n) of the l-th prime in the primefactorization of n ∈ N.
3. The residue res(a,n) of a modulo n.
4. Restricted subtraction a ·− b.
DMACC John M. Gillespie 11
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Examples
I If we let
S(n) =n∑
i=1
s̄g(res(n, i))
then S(n) yields the number of divisors of n.
I If we let
π(n) =n∑
i=2
s̄g(S(n) ·− 2)
then π(n) yields the number of primes among 2, ..., n.
Then the nth prime is the least j such that π(j) = n + 1.
DMACC John M. Gillespie 12
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Examples
I If we let
S(n) =n∑
i=1
s̄g(res(n, i))
then S(n) yields the number of divisors of n.
I If we let
π(n) =n∑
i=2
s̄g(S(n) ·− 2)
then π(n) yields the number of primes among 2, ..., n.
Then the nth prime is the least j such that π(j) = n + 1.
DMACC John M. Gillespie 13
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Examples
I If we let
S(n) =n∑
i=1
s̄g(res(n, i))
then S(n) yields the number of divisors of n.
I If we let
π(n) =n∑
i=2
s̄g(S(n) ·− 2)
then π(n) yields the number of primes among 2, ..., n.
Then the nth prime is the least j such that π(j) = n + 1.
DMACC John M. Gillespie 14
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Examples
I If we let
S(n) =n∑
i=1
s̄g(res(n, i))
then S(n) yields the number of divisors of n.
I If we let
π(n) =n∑
i=2
s̄g(S(n) ·− 2)
then π(n) yields the number of primes among 2, ..., n.
Then the nth prime is the least j such that π(j) = n + 1.
DMACC John M. Gillespie 15
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Examples
I If we let
S(n) =n∑
i=1
s̄g(res(n, i))
then S(n) yields the number of divisors of n.
I If we let
π(n) =n∑
i=2
s̄g(S(n) ·− 2)
then π(n) yields the number of primes among 2, ..., n.
Then the nth prime is the least j such that π(j) = n + 1.
DMACC John M. Gillespie 16
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Bounded search
Suppose a function τ(a1, ..., ar , n) is given which satisfies thefollowing formula.
(∀a1 . . . ∀ar∃n)(τ(~a, n) = 0)
Then the smallest n for which τ(a1, ..., ar , n) = 0 is given by theexpression below.
B∑i=0
i∏j=0
τ(~a, j)sg
= τ(~a, 0)sg + τ(~a, 0)sgτ(~a, 1)sg + τ(~a, 0)sgτ(~a, 1)sgτ(~a, 2)sg + . . .
+ τ(~a, 0)sgτ(~a, 1)sg . . . τ(~a,B)sg
DMACC John M. Gillespie 17
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Bounded search
Suppose a function τ(a1, ..., ar , n) is given which satisfies thefollowing formula.
(∀a1 . . . ∀ar∃n)(τ(~a, n) = 0)
Then the smallest n for which τ(a1, ..., ar , n) = 0 is given by theexpression below.
B∑i=0
i∏j=0
τ(~a, j)sg
= τ(~a, 0)sg + τ(~a, 0)sgτ(~a, 1)sg + τ(~a, 0)sgτ(~a, 1)sgτ(~a, 2)sg + . . .
+ τ(~a, 0)sgτ(~a, 1)sg . . . τ(~a,B)sg
DMACC John M. Gillespie 18
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Bounded search
Suppose a function τ(a1, ..., ar , n) is given which satisfies thefollowing formula.
(∀a1 . . . ∀ar∃n)(τ(~a, n) = 0)
Then the smallest n for which τ(a1, ..., ar , n) = 0 is given by theexpression below.
B∑i=0
i∏j=0
τ(~a, j)sg
= τ(~a, 0)sg + τ(~a, 0)sgτ(~a, 1)sg + τ(~a, 0)sgτ(~a, 1)sgτ(~a, 2)sg + . . .
+ τ(~a, 0)sgτ(~a, 1)sg . . . τ(~a,B)sg
DMACC John M. Gillespie 19
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Bounded search
The following notation is fairly standard.
µj [τ(~a, j) = 0] =B∑i=0
i∏j=0
τ(~a, j)sg
I The bound B is specific to the function τ .
I For example the nth prime pn satisfies pn < 22n .
I This provides the bound B = 22n .
DMACC John M. Gillespie 20
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Bounded search
The following notation is fairly standard.
µj [τ(~a, j) = 0] =B∑i=0
i∏j=0
τ(~a, j)sg
I The bound B is specific to the function τ .
I For example the nth prime pn satisfies pn < 22n .
I This provides the bound B = 22n .
DMACC John M. Gillespie 21
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Bounded search
The following notation is fairly standard.
µj [τ(~a, j) = 0] =B∑i=0
i∏j=0
τ(~a, j)sg
I The bound B is specific to the function τ .
I For example the nth prime pn satisfies pn < 22n .
I This provides the bound B = 22n .
DMACC John M. Gillespie 22
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Bounded search
The following notation is fairly standard.
µj [τ(~a, j) = 0] =B∑i=0
i∏j=0
τ(~a, j)sg
I The bound B is specific to the function τ .
I For example the nth prime pn satisfies pn < 22n .
I This provides the bound B = 22n .
DMACC John M. Gillespie 23
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The length of n ∈ N
Consider the problem of finding the index l of the largest primedividing n ∈ N and suppose n > 1.
I Take the sign of the sum
sg(expl(n) + exp2(n) + . . .+ expn(n)).
I What we seek is the smallest j such that
sg(expj+1(n) + expj+2(n) + . . .+ expn(n)) = 0.
DMACC John M. Gillespie 24
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The length of n ∈ N
Consider the problem of finding the index l of the largest primedividing n ∈ N and suppose n > 1.
I Take the sign of the sum
sg(expl(n) + exp2(n) + . . .+ expn(n)).
I What we seek is the smallest j such that
sg(expj+1(n) + expj+2(n) + . . .+ expn(n)) = 0.
DMACC John M. Gillespie 25
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The length of n ∈ N
Consider the problem of finding the index l of the largest primedividing n ∈ N and suppose n > 1.
I Take the sign of the sum
sg(expl(n) + exp2(n) + . . .+ expn(n)).
I What we seek is the smallest j such that
sg(expj+1(n) + expj+2(n) + . . .+ expn(n)) = 0.
DMACC John M. Gillespie 26
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The length of n ∈ N
The following primitive recursive function yields the index of thelargest prime divisor of the natural number n.
long(n) =n∑
k=0
k∏j=0
sg
n∑l=j+1
expl(n)
I The bound n is sufficient since n < pn.
I This yields the smallest j such that expl(n) = 0 if l > j .
DMACC John M. Gillespie 27
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The length of n ∈ N
The following primitive recursive function yields the index of thelargest prime divisor of the natural number n.
long(n) =n∑
k=0
k∏j=0
sg
n∑l=j+1
expl(n)
I The bound n is sufficient since n < pn.
I This yields the smallest j such that expl(n) = 0 if l > j .
DMACC John M. Gillespie 28
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The length of n ∈ N
The following primitive recursive function yields the index of thelargest prime divisor of the natural number n.
long(n) =n∑
k=0
k∏j=0
sg
n∑l=j+1
expl(n)
I The bound n is sufficient since n < pn.
I This yields the smallest j such that expl(n) = 0 if l > j .
DMACC John M. Gillespie 29
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Other forms of recursion
Course-of-values recursion
DefinitionGiven a binary function β and any a ∈ N define ϕ as follows;
ϕ(0) = a and ϕ(n + 1) = β(n∏
i=0
pϕ(i)i , n).
We can prove that such a definition yields a primitive recursivefunction.
DMACC John M. Gillespie 30
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Course-of-values recursion
Definition of ϕ
ϕ(0) = a and ϕ(n + 1) = β(n∏
i=0
pϕ(i)i , n)
Define a function ψ(n) =∏n
i=0 pϕ(n)i . Then ψ(0) = 2a and
ψ(n + 1) = ψ(n) · pϕ(n+1)n+1
= ψ(n) · pβ(ψ(n),n)n+1
Thus ψ is primitive recursive and ϕ(n) = expn ψ(n) hence ϕ is alsoprimitive recursive.
DMACC John M. Gillespie 31
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Course-of-values recursion
Definition of ϕ
ϕ(0) = a and ϕ(n + 1) = β(n∏
i=0
pϕ(i)i , n)
Define a function ψ(n) =∏n
i=0 pϕ(n)i . Then ψ(0) = 2a and
ψ(n + 1) = ψ(n) · pϕ(n+1)n+1
= ψ(n) · pβ(ψ(n),n)n+1
Thus ψ is primitive recursive and ϕ(n) = expn ψ(n) hence ϕ is alsoprimitive recursive.
DMACC John M. Gillespie 32
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Course-of-values recursion
Definition of ϕ
ϕ(0) = a and ϕ(n + 1) = β(n∏
i=0
pϕ(i)i , n)
Define a function ψ(n) =∏n
i=0 pϕ(n)i . Then ψ(0) = 2a and
ψ(n + 1) = ψ(n) · pϕ(n+1)n+1
= ψ(n) · pβ(ψ(n),n)n+1
Thus ψ is primitive recursive and ϕ(n) = expn ψ(n) hence ϕ is alsoprimitive recursive.
DMACC John M. Gillespie 33
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Course-of-values recursion
Definition of ϕ
ϕ(0) = a and ϕ(n + 1) = β(n∏
i=0
pϕ(i)i , n)
Define a function ψ(n) =∏n
i=0 pϕ(n)i . Then ψ(0) = 2a and
ψ(n + 1) = ψ(n) · pϕ(n+1)n+1
= ψ(n) · pβ(ψ(n),n)n+1
Thus ψ is primitive recursive and ϕ(n) = expn ψ(n) hence ϕ is alsoprimitive recursive.
DMACC John M. Gillespie 34
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Another form of recursion
Simultaneous recursion
DefinitionGiven a, b ∈ N and binary functions β1 and β2 define σ1 and σ2;σ1(0) = a, σ2(0) = b with
σ1(n + 1) = β1(σ1(n), σ2(n))
σ2(n + 1) = β2(σ1(n), σ2(n)).
We can prove that such a definition yields primitive recursivefunctins σ1 and σ2.
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Another form of recursion
Simultaneous recursion
DefinitionGiven a, b ∈ N and binary functions β1 and β2 define σ1 and σ2;σ1(0) = a, σ2(0) = b with
σ1(n + 1) = β1(σ1(n), σ2(n))
σ2(n + 1) = β2(σ1(n), σ2(n)).
We can prove that such a definition yields primitive recursivefunctins σ1 and σ2.
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Simultaneous recursion
σ1(0) = a and σ2(0) = b
σ1(n + 1) = β1(σ1(n), σ2(n))
σ2(n + 1) = β2(σ1(n), σ2(n)).
Define a function ψ(n) = pσ1(n)1 ·pσ2(n)
2 . So then ψ(0) = pa1 ·pb2 and
ψ(n + 1) = pσ1(n+1)1 · pσ2(n+1)
2
= pβ1(σ1(n),σ2(n))1 · pβ2(σ1(n),σ2(n)
2
= pβ1(exp1 ψ(n),exp2 ψ(n))1 · pβ2(exp1 ψ(n),exp2 ψ(n))
2
Hence ψ is primitive recursive ⇒ σ1 and σ2 are primitive recursive.
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Simultaneous recursion
σ1(0) = a and σ2(0) = b
σ1(n + 1) = β1(σ1(n), σ2(n))
σ2(n + 1) = β2(σ1(n), σ2(n)).
Define a function ψ(n) = pσ1(n)1 ·pσ2(n)
2 . So then ψ(0) = pa1 ·pb2 and
ψ(n + 1) = pσ1(n+1)1 · pσ2(n+1)
2
= pβ1(σ1(n),σ2(n))1 · pβ2(σ1(n),σ2(n)
2
= pβ1(exp1 ψ(n),exp2 ψ(n))1 · pβ2(exp1 ψ(n),exp2 ψ(n))
2
Hence ψ is primitive recursive ⇒ σ1 and σ2 are primitive recursive.
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Simultaneous recursion
σ1(0) = a and σ2(0) = b
σ1(n + 1) = β1(σ1(n), σ2(n))
σ2(n + 1) = β2(σ1(n), σ2(n)).
Define a function ψ(n) = pσ1(n)1 ·pσ2(n)
2 . So then ψ(0) = pa1 ·pb2 and
ψ(n + 1) = pσ1(n+1)1 · pσ2(n+1)
2
= pβ1(σ1(n),σ2(n))1 · pβ2(σ1(n),σ2(n)
2
= pβ1(exp1 ψ(n),exp2 ψ(n))1 · pβ2(exp1 ψ(n),exp2 ψ(n))
2
Hence ψ is primitive recursive ⇒ σ1 and σ2 are primitive recursive.
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Simultaneous recursion
σ1(0) = a and σ2(0) = b
σ1(n + 1) = β1(σ1(n), σ2(n))
σ2(n + 1) = β2(σ1(n), σ2(n)).
Define a function ψ(n) = pσ1(n)1 ·pσ2(n)
2 . So then ψ(0) = pa1 ·pb2 and
ψ(n + 1) = pσ1(n+1)1 · pσ2(n+1)
2
= pβ1(σ1(n),σ2(n))1 · pβ2(σ1(n),σ2(n)
2
= pβ1(exp1 ψ(n),exp2 ψ(n))1 · pβ2(exp1 ψ(n),exp2 ψ(n))
2
Hence ψ is primitive recursive ⇒ σ1 and σ2 are primitive recursive.
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Nested recursion
Example
Given known functions α, β and γ define ϕ(0, a) = α(a) with
ϕ(n + 1, a) = β(n, ϕ(n, γ(n, a, ϕ(n, a)))).
I This function is indeed primitive recursive.
I But in such a definition if you have induction on multiplevariables then your function may no longer be PR.
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Nested recursion
Example
Given known functions α, β and γ define ϕ(0, a) = α(a) with
ϕ(n + 1, a) = β(n, ϕ(n, γ(n, a, ϕ(n, a)))).
I This function is indeed primitive recursive.
I But in such a definition if you have induction on multiplevariables then your function may no longer be PR.
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Nested recursion
Example
Given known functions α, β and γ define ϕ(0, a) = α(a) with
ϕ(n + 1, a) = β(n, ϕ(n, γ(n, a, ϕ(n, a)))).
I This function is indeed primitive recursive.
I But in such a definition if you have induction on multiplevariables then your function may no longer be PR.
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General recursive functions
General recursive functions are defined in terms of a system ofequations.
I The class of general recursive functions coincides with theclass of all computable functions.
I Hence functions of the λ-calculus, Turing computablefunctions, and so on are all general recursive functions.
I Given a defining system of equations you may not be able totell if you have a working definition.
I In general the problem is undecidable.
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General recursive functions
General recursive functions are defined in terms of a system ofequations.
I The class of general recursive functions coincides with theclass of all computable functions.
I Hence functions of the λ-calculus, Turing computablefunctions, and so on are all general recursive functions.
I Given a defining system of equations you may not be able totell if you have a working definition.
I In general the problem is undecidable.
DMACC John M. Gillespie 45
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General recursive functions
General recursive functions are defined in terms of a system ofequations.
I The class of general recursive functions coincides with theclass of all computable functions.
I Hence functions of the λ-calculus, Turing computablefunctions, and so on are all general recursive functions.
I Given a defining system of equations you may not be able totell if you have a working definition.
I In general the problem is undecidable.
DMACC John M. Gillespie 46
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General recursive functions
General recursive functions are defined in terms of a system ofequations.
I The class of general recursive functions coincides with theclass of all computable functions.
I Hence functions of the λ-calculus, Turing computablefunctions, and so on are all general recursive functions.
I Given a defining system of equations you may not be able totell if you have a working definition.
I In general the problem is undecidable.
DMACC John M. Gillespie 47
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General recursive functions
General recursive functions are defined in terms of a system ofequations.
I The class of general recursive functions coincides with theclass of all computable functions.
I Hence functions of the λ-calculus, Turing computablefunctions, and so on are all general recursive functions.
I Given a defining system of equations you may not be able totell if you have a working definition.
I In general the problem is undecidable.
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General recursive functions
Given a defining system of equations there are two numbers whichare noteworthy.
The index of the function being defined σs and the number Aof axioms.
Example
1. σ1(0, x2) = x2
2. σ1(σ0(x1), x2) = σ0(σ1(x1, x2))
3. σ2(0, x2) = 0
4. σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2)
5. σ3(x1) = σ2(x1, x1)
I For this example s = 3, A = 5 and σ3 is the square function.
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General recursive functions
Given a defining system of equations there are two numbers whichare noteworthy.
The index of the function being defined σs and the number Aof axioms.
Example
1. σ1(0, x2) = x2
2. σ1(σ0(x1), x2) = σ0(σ1(x1, x2))
3. σ2(0, x2) = 0
4. σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2)
5. σ3(x1) = σ2(x1, x1)
I For this example s = 3, A = 5 and σ3 is the square function.
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General recursive functions
Given a defining system of equations there are two numbers whichare noteworthy.
The index of the function being defined σs and the number Aof axioms.
Example
1. σ1(0, x2) = x2
2. σ1(σ0(x1), x2) = σ0(σ1(x1, x2))
3. σ2(0, x2) = 0
4. σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2)
5. σ3(x1) = σ2(x1, x1)
I For this example s = 3, A = 5 and σ3 is the square function.
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General recursive functions
Given a defining system of equations there are two numbers whichare noteworthy.
The index of the function being defined σs and the number Aof axioms.
Example
1. σ1(0, x2) = x2
2. σ1(σ0(x1), x2) = σ0(σ1(x1, x2))
3. σ2(0, x2) = 0
4. σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2)
5. σ3(x1) = σ2(x1, x1)
I For this example s = 3, A = 5 and σ3 is the square function.
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General recursive functions
Example
1. σ1(0) = 0
2. σ1(σ0(x1)) = x1
3. σ2(0, x2) = x2
4. σ2(σ0(x1), x2) = σ0(σ2(x1, x2))
5. σ3(0) = 1
6. σ3(σ0(x1)) = σ2(ϕ(x1), ϕ(σ1(x1)))
I For this example s = 3, A = 6. Can you identify the functionσ3?
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General recursive functions
Example
1. σ1(0) = 0
2. σ1(σ0(x1)) = x1
3. σ2(0, x2) = x2
4. σ2(σ0(x1), x2) = σ0(σ2(x1, x2))
5. σ3(0) = 1
6. σ3(σ0(x1)) = σ2(ϕ(x1), ϕ(σ1(x1)))
I For this example s = 3, A = 6. Can you identify the functionσ3?
DMACC John M. Gillespie 54
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General recursive functions
Example
1. σ1(0) = 0
2. σ1(σ0(x1)) = x1
3. σ2(x1, 0) = x1
4. σ2(x1, σ0(x2)) = σ1(σ2(x1, x2))
I For this example s = 2 and A = 4.
I σ2(x1, x2) = x1 ·− x2.
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General recursive functions
Example
1. σ1(0) = 0
2. σ1(σ0(x1)) = x1
3. σ2(x1, 0) = x1
4. σ2(x1, σ0(x2)) = σ1(σ2(x1, x2))
I For this example s = 2 and A = 4.
I σ2(x1, x2) = x1 ·− x2.
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Computations as deductions
Axioms
{σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)
Deductions
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (3)z(3, 2) σ1(0, σ0(0)) = σ0(0) (4)n(2, 2) σ1(σ0(x1), σ0(x2)) = σ0(σ1(x1, σ0(x2))) (5)z(5, i) σ1(σ0(0), σ0(0)) = σ0(σ1(0, σ0(0))) (6)sub(4, 6, 15) σ1(σ0(0), σ0(0)) = σ0(σ0(0)) (7)
This illustrates the three admissable steps in any computation;n, z and sub.
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Computations as deductions
Axioms
{σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)
Deductions
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (3)
z(3, 2) σ1(0, σ0(0)) = σ0(0) (4)n(2, 2) σ1(σ0(x1), σ0(x2)) = σ0(σ1(x1, σ0(x2))) (5)z(5, i) σ1(σ0(0), σ0(0)) = σ0(σ1(0, σ0(0))) (6)sub(4, 6, 15) σ1(σ0(0), σ0(0)) = σ0(σ0(0)) (7)
This illustrates the three admissable steps in any computation;n, z and sub.
DMACC John M. Gillespie 58
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Computations as deductions
Axioms
{σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)
Deductions
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (3)z(3, 2) σ1(0, σ0(0)) = σ0(0) (4)
n(2, 2) σ1(σ0(x1), σ0(x2)) = σ0(σ1(x1, σ0(x2))) (5)z(5, i) σ1(σ0(0), σ0(0)) = σ0(σ1(0, σ0(0))) (6)sub(4, 6, 15) σ1(σ0(0), σ0(0)) = σ0(σ0(0)) (7)
This illustrates the three admissable steps in any computation;n, z and sub.
DMACC John M. Gillespie 59
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Computations as deductions
Axioms
{σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)
Deductions
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (3)z(3, 2) σ1(0, σ0(0)) = σ0(0) (4)n(2, 2) σ1(σ0(x1), σ0(x2)) = σ0(σ1(x1, σ0(x2))) (5)
z(5, i) σ1(σ0(0), σ0(0)) = σ0(σ1(0, σ0(0))) (6)sub(4, 6, 15) σ1(σ0(0), σ0(0)) = σ0(σ0(0)) (7)
This illustrates the three admissable steps in any computation;n, z and sub.
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Computations as deductions
Axioms
{σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)
Deductions
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (3)z(3, 2) σ1(0, σ0(0)) = σ0(0) (4)n(2, 2) σ1(σ0(x1), σ0(x2)) = σ0(σ1(x1, σ0(x2))) (5)z(5, i) σ1(σ0(0), σ0(0)) = σ0(σ1(0, σ0(0))) (6)
sub(4, 6, 15) σ1(σ0(0), σ0(0)) = σ0(σ0(0)) (7)
This illustrates the three admissable steps in any computation;n, z and sub.
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Computations as deductions
Axioms
{σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)
Deductions
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (3)z(3, 2) σ1(0, σ0(0)) = σ0(0) (4)n(2, 2) σ1(σ0(x1), σ0(x2)) = σ0(σ1(x1, σ0(x2))) (5)z(5, i) σ1(σ0(0), σ0(0)) = σ0(σ1(0, σ0(0))) (6)sub(4, 6, 15) σ1(σ0(0), σ0(0)) = σ0(σ0(0)) (7)
This illustrates the three admissable steps in any computation;n, z and sub.
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Computations as deductions
Axioms
{σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)
Deductions
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (3)z(3, 2) σ1(0, σ0(0)) = σ0(0) (4)n(2, 2) σ1(σ0(x1), σ0(x2)) = σ0(σ1(x1, σ0(x2))) (5)z(5, i) σ1(σ0(0), σ0(0)) = σ0(σ1(0, σ0(0))) (6)sub(4, 6, 15) σ1(σ0(0), σ0(0)) = σ0(σ0(0)) (7)
This illustrates the three admissable steps in any computation;n, z and sub.
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The admissable steps
I z(i , j) denotes plugging zero into the j variable of equation i .
I n(i , j) denotes non-zero evaluation; We are plugging σ0(xj)into equation i in the variable j .
I sub(i , j , k) denotes substitution of RHS of equation i intoequation j at the kth position.
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The admissable steps
I z(i , j) denotes plugging zero into the j variable of equation i .
I n(i , j) denotes non-zero evaluation; We are plugging σ0(xj)into equation i in the variable j .
I sub(i , j , k) denotes substitution of RHS of equation i intoequation j at the kth position.
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The admissable steps
I z(i , j) denotes plugging zero into the j variable of equation i .
I n(i , j) denotes non-zero evaluation; We are plugging σ0(xj)into equation i in the variable j .
I sub(i , j , k) denotes substitution of RHS of equation i intoequation j at the kth position.
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Computations as deductions
Axioms
σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)σ2(0, x2) = 0 (3)σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2) (4)
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (5)z(5, 2) σ1(0, σ0(0)) = σ0(0) (6)n(4, 2) σ2(σ0(x1), σ0(x2)) = σ1(σ2(x1, σ0(x2)), σ0(x2)) (7)z(7, i) σ2(σ0(0), σ0(0)) = σ1(σ2(0, σ0(0)), σ0(0)) (8)n(3, 2) σ2(0, σ0(x2)) = 0 (9)z(9, 2) σ2(0, σ0(0)) = 0 (10)sub(10, 8) σ2(σ0(0), σ0(0))) = σ1(0, σ0(0)) (11)sub(6, 11) σ2(σ0(0), σ0(0))) = σ0(0) (12)
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Computations as deductions
Axioms
σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)σ2(0, x2) = 0 (3)σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2) (4)
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (5)
z(5, 2) σ1(0, σ0(0)) = σ0(0) (6)n(4, 2) σ2(σ0(x1), σ0(x2)) = σ1(σ2(x1, σ0(x2)), σ0(x2)) (7)z(7, i) σ2(σ0(0), σ0(0)) = σ1(σ2(0, σ0(0)), σ0(0)) (8)n(3, 2) σ2(0, σ0(x2)) = 0 (9)z(9, 2) σ2(0, σ0(0)) = 0 (10)sub(10, 8) σ2(σ0(0), σ0(0))) = σ1(0, σ0(0)) (11)sub(6, 11) σ2(σ0(0), σ0(0))) = σ0(0) (12)
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Computations as deductions
Axioms
σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)σ2(0, x2) = 0 (3)σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2) (4)
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (5)z(5, 2) σ1(0, σ0(0)) = σ0(0) (6)
n(4, 2) σ2(σ0(x1), σ0(x2)) = σ1(σ2(x1, σ0(x2)), σ0(x2)) (7)z(7, i) σ2(σ0(0), σ0(0)) = σ1(σ2(0, σ0(0)), σ0(0)) (8)n(3, 2) σ2(0, σ0(x2)) = 0 (9)z(9, 2) σ2(0, σ0(0)) = 0 (10)sub(10, 8) σ2(σ0(0), σ0(0))) = σ1(0, σ0(0)) (11)sub(6, 11) σ2(σ0(0), σ0(0))) = σ0(0) (12)
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Computations as deductions
Axioms
σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)σ2(0, x2) = 0 (3)σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2) (4)
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (5)z(5, 2) σ1(0, σ0(0)) = σ0(0) (6)n(4, 2) σ2(σ0(x1), σ0(x2)) = σ1(σ2(x1, σ0(x2)), σ0(x2)) (7)
z(7, i) σ2(σ0(0), σ0(0)) = σ1(σ2(0, σ0(0)), σ0(0)) (8)n(3, 2) σ2(0, σ0(x2)) = 0 (9)z(9, 2) σ2(0, σ0(0)) = 0 (10)sub(10, 8) σ2(σ0(0), σ0(0))) = σ1(0, σ0(0)) (11)sub(6, 11) σ2(σ0(0), σ0(0))) = σ0(0) (12)
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Computations as deductions
Axioms
σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)σ2(0, x2) = 0 (3)σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2) (4)
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (5)z(5, 2) σ1(0, σ0(0)) = σ0(0) (6)n(4, 2) σ2(σ0(x1), σ0(x2)) = σ1(σ2(x1, σ0(x2)), σ0(x2)) (7)z(7, i) σ2(σ0(0), σ0(0)) = σ1(σ2(0, σ0(0)), σ0(0)) (8)
n(3, 2) σ2(0, σ0(x2)) = 0 (9)z(9, 2) σ2(0, σ0(0)) = 0 (10)sub(10, 8) σ2(σ0(0), σ0(0))) = σ1(0, σ0(0)) (11)sub(6, 11) σ2(σ0(0), σ0(0))) = σ0(0) (12)
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Computations as deductions
Axioms
σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)σ2(0, x2) = 0 (3)σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2) (4)
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (5)z(5, 2) σ1(0, σ0(0)) = σ0(0) (6)n(4, 2) σ2(σ0(x1), σ0(x2)) = σ1(σ2(x1, σ0(x2)), σ0(x2)) (7)z(7, i) σ2(σ0(0), σ0(0)) = σ1(σ2(0, σ0(0)), σ0(0)) (8)n(3, 2) σ2(0, σ0(x2)) = 0 (9)
z(9, 2) σ2(0, σ0(0)) = 0 (10)sub(10, 8) σ2(σ0(0), σ0(0))) = σ1(0, σ0(0)) (11)sub(6, 11) σ2(σ0(0), σ0(0))) = σ0(0) (12)
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Computations as deductions
Axioms
σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)σ2(0, x2) = 0 (3)σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2) (4)
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (5)z(5, 2) σ1(0, σ0(0)) = σ0(0) (6)n(4, 2) σ2(σ0(x1), σ0(x2)) = σ1(σ2(x1, σ0(x2)), σ0(x2)) (7)z(7, i) σ2(σ0(0), σ0(0)) = σ1(σ2(0, σ0(0)), σ0(0)) (8)n(3, 2) σ2(0, σ0(x2)) = 0 (9)z(9, 2) σ2(0, σ0(0)) = 0 (10)
sub(10, 8) σ2(σ0(0), σ0(0))) = σ1(0, σ0(0)) (11)sub(6, 11) σ2(σ0(0), σ0(0))) = σ0(0) (12)
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Computations as deductions
Axioms
σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)σ2(0, x2) = 0 (3)σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2) (4)
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (5)z(5, 2) σ1(0, σ0(0)) = σ0(0) (6)n(4, 2) σ2(σ0(x1), σ0(x2)) = σ1(σ2(x1, σ0(x2)), σ0(x2)) (7)z(7, i) σ2(σ0(0), σ0(0)) = σ1(σ2(0, σ0(0)), σ0(0)) (8)n(3, 2) σ2(0, σ0(x2)) = 0 (9)z(9, 2) σ2(0, σ0(0)) = 0 (10)sub(10, 8) σ2(σ0(0), σ0(0))) = σ1(0, σ0(0)) (11)
sub(6, 11) σ2(σ0(0), σ0(0))) = σ0(0) (12)
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Computations as deductions
Axioms
σ1(0, x2) = x2 (1)σ1(σ0(x1), x2) = σ0(σ1(x1, x2)) (2)σ2(0, x2) = 0 (3)σ2(σ0(x1), x2) = σ1(σ2(x1, x2), x2) (4)
n(1, 2) σ1(0, σ0(x2)) = σ0(x2) (5)z(5, 2) σ1(0, σ0(0)) = σ0(0) (6)n(4, 2) σ2(σ0(x1), σ0(x2)) = σ1(σ2(x1, σ0(x2)), σ0(x2)) (7)z(7, i) σ2(σ0(0), σ0(0)) = σ1(σ2(0, σ0(0)), σ0(0)) (8)n(3, 2) σ2(0, σ0(x2)) = 0 (9)z(9, 2) σ2(0, σ0(0)) = 0 (10)sub(10, 8) σ2(σ0(0), σ0(0))) = σ1(0, σ0(0)) (11)sub(6, 11) σ2(σ0(0), σ0(0))) = σ0(0) (12)
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Key component of the Kleene normal form
The Kleene operator
Given a function τ(a1, ..., ar , n) in which for every ~a there exists ann such that τ(~a, n) = 0 we can define a new function ϕ(~a).
ϕ(a1, ..., ar ) = µj [τ(a1, ..., ar ) = 0]
This is the unbounded search.
limB→∞
B∑i=0
i∏j=0
τ(~a, j)sg
We will use this Kleene operator later to search throughcomputations.
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Key component of the Kleene normal form
The Kleene operator
Given a function τ(a1, ..., ar , n) in which for every ~a there exists ann such that τ(~a, n) = 0 we can define a new function ϕ(~a).
ϕ(a1, ..., ar ) = µj [τ(a1, ..., ar ) = 0]
This is the unbounded search.
limB→∞
B∑i=0
i∏j=0
τ(~a, j)sg
We will use this Kleene operator later to search throughcomputations.
DMACC John M. Gillespie 77
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Key component of the Kleene normal form
The Kleene operator
Given a function τ(a1, ..., ar , n) in which for every ~a there exists ann such that τ(~a, n) = 0 we can define a new function ϕ(~a).
ϕ(a1, ..., ar ) = µj [τ(a1, ..., ar ) = 0]
This is the unbounded search.
limB→∞
B∑i=0
i∏j=0
τ(~a, j)sg
We will use this Kleene operator later to search throughcomputations.
DMACC John M. Gillespie 78
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Key component of the Kleene normal form
The Kleene operator
Given a function τ(a1, ..., ar , n) in which for every ~a there exists ann such that τ(~a, n) = 0 we can define a new function ϕ(~a).
ϕ(a1, ..., ar ) = µj [τ(a1, ..., ar ) = 0]
This is the unbounded search.
limB→∞
B∑i=0
i∏j=0
τ(~a, j)sg
We will use this Kleene operator later to search throughcomputations.
DMACC John M. Gillespie 79
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The Kleene operator as a general recursive function
Suppose a function τ(a1, ..., ar , n) is already given. We define anew function σ.
I σ(j , a1, ..., ar , 0) = j
I σ(j , a1, ..., ar , n + 1) = σ(j + 1, a1, ..., ar , τ(a1, ..., ar , j + 1))
Then µj [τ(a1, ..., ar , j) = 0] = σ(0, a1, ..., ar , τ(a1, ..., ar , 0)).
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The Kleene operator as a general recursive function
Suppose a function τ(a1, ..., ar , n) is already given. We define anew function σ.
I σ(j , a1, ..., ar , 0) = j
I σ(j , a1, ..., ar , n + 1) = σ(j + 1, a1, ..., ar , τ(a1, ..., ar , j + 1))
Then µj [τ(a1, ..., ar , j) = 0] = σ(0, a1, ..., ar , τ(a1, ..., ar , 0)).
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Kleene’s theorem states that any general recursive functionϕ can be put in the form
ϕ = ψ(µ[τ ])
where µ is the Kleene operator and ψ and τ are primitiverecursive.
I This is the normal form we wish to prove exists.
I This proves the Kleene operator cannot be primitive recursive.
I This also proves that if we augment the class PR with theKleene operator this yields the class of all computablefunctions.
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Kleene’s theorem states that any general recursive functionϕ can be put in the form
ϕ = ψ(µ[τ ])
where µ is the Kleene operator and ψ and τ are primitiverecursive.
I This is the normal form we wish to prove exists.
I This proves the Kleene operator cannot be primitive recursive.
I This also proves that if we augment the class PR with theKleene operator this yields the class of all computablefunctions.
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Kleene’s theorem states that any general recursive functionϕ can be put in the form
ϕ = ψ(µ[τ ])
where µ is the Kleene operator and ψ and τ are primitiverecursive.
I This is the normal form we wish to prove exists.
I This proves the Kleene operator cannot be primitive recursive.
I This also proves that if we augment the class PR with theKleene operator this yields the class of all computablefunctions.
DMACC John M. Gillespie 84
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Kleene’s theorem states that any general recursive functionϕ can be put in the form
ϕ = ψ(µ[τ ])
where µ is the Kleene operator and ψ and τ are primitiverecursive.
I This is the normal form we wish to prove exists.
I This proves the Kleene operator cannot be primitive recursive.
I This also proves that if we augment the class PR with theKleene operator this yields the class of all computablefunctions.
DMACC John M. Gillespie 85
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Arithmetization of computations
We begin with the basic Godel numbering of symbols.
Individual symbol Godel numberσn 2n + 2, n = 0, 1, ...0 1= 3( 5) 7, 9xn 2n + 9, n = 1, 2, ...
By Godel numbering the symbols used in computations we canemploy standard arithmetic to sort through sets of computations.Furthermore we can hone in on those computations which areactually correct.
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Arithmetization of computations
We begin with the basic Godel numbering of symbols.
Individual symbol Godel numberσn 2n + 2, n = 0, 1, ...0 1= 3( 5) 7, 9xn 2n + 9, n = 1, 2, ...
By Godel numbering the symbols used in computations we canemploy standard arithmetic to sort through sets of computations.
Furthermore we can hone in on those computations which areactually correct.
DMACC John M. Gillespie 87
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Arithmetization of computations
We begin with the basic Godel numbering of symbols.
Individual symbol Godel numberσn 2n + 2, n = 0, 1, ...0 1= 3( 5) 7, 9xn 2n + 9, n = 1, 2, ...
By Godel numbering the symbols used in computations we canemploy standard arithmetic to sort through sets of computations.Furthermore we can hone in on those computations which areactually correct.
DMACC John M. Gillespie 88
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More on Godel numbers
Some examples of symbol sequences and their Godel numbers.
Example
1. σ0(0) has Godel number p21p
52p
13p
74
2. σ0(σ0(0)) has Godel number p21p
52p
23p
54p
15p
76p
77
3. σ2(x1) = x1 has Godel number p41p
52p
113 p7
4p35p
116
4. σ0(xi ) has Godel number p21p
52p
2i+93 p7
4
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More on Godel numbers
Some examples of symbol sequences and their Godel numbers.
Example
1. σ0(0) has Godel number p21p
52p
13p
74
2. σ0(σ0(0)) has Godel number p21p
52p
23p
54p
15p
76p
77
3. σ2(x1) = x1 has Godel number p41p
52p
113 p7
4p35p
116
4. σ0(xi ) has Godel number p21p
52p
2i+93 p7
4
DMACC John M. Gillespie 90
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More on Godel numbers
Some examples of symbol sequences and their Godel numbers.
Example
1. σ0(0) has Godel number p21p
52p
13p
74
2. σ0(σ0(0)) has Godel number p21p
52p
23p
54p
15p
76p
77
3. σ2(x1) = x1 has Godel number p41p
52p
113 p7
4p35p
116
4. σ0(xi ) has Godel number p21p
52p
2i+93 p7
4
DMACC John M. Gillespie 91
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More on Godel numbers
Some examples of symbol sequences and their Godel numbers.
Example
1. σ0(0) has Godel number p21p
52p
13p
74
2. σ0(σ0(0)) has Godel number p21p
52p
23p
54p
15p
76p
77
3. σ2(x1) = x1 has Godel number p41p
52p
113 p7
4p35p
116
4. σ0(xi ) has Godel number p21p
52p
2i+93 p7
4
DMACC John M. Gillespie 92
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Godel numbering a computation
This is the first example computation.
Axioms
{q1 := p4
1p52p
13p
94p
135 p7
6p37p
138
. . .
Deductions
q3 := p4
1p52 . . . p
1313p
714
...q7 := p4
1p52 . . . p
719p
720
The entire computation is encoded in the Godel numberpq1
1 pq22 . . . pq7
7 .
DMACC John M. Gillespie 93
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Basic strategy
I Every computation is assigned a Godel number pe11 pe2
2 . . . pell .
I Each exponent ei is the Godel number of an equation.
I The first A equations e1,...,eA are the axioms.
I The rest eA+1,...,el are expected to be deductions fromprevious equations.
I We require functions which can detect such deductions.
DMACC John M. Gillespie 94
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Basic strategy
I Every computation is assigned a Godel number pe11 pe2
2 . . . pell .
I Each exponent ei is the Godel number of an equation.
I The first A equations e1,...,eA are the axioms.
I The rest eA+1,...,el are expected to be deductions fromprevious equations.
I We require functions which can detect such deductions.
DMACC John M. Gillespie 95
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Basic strategy
I Every computation is assigned a Godel number pe11 pe2
2 . . . pell .
I Each exponent ei is the Godel number of an equation.
I The first A equations e1,...,eA are the axioms.
I The rest eA+1,...,el are expected to be deductions fromprevious equations.
I We require functions which can detect such deductions.
DMACC John M. Gillespie 96
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Basic strategy
I Every computation is assigned a Godel number pe11 pe2
2 . . . pell .
I Each exponent ei is the Godel number of an equation.
I The first A equations e1,...,eA are the axioms.
I The rest eA+1,...,el are expected to be deductions fromprevious equations.
I We require functions which can detect such deductions.
DMACC John M. Gillespie 97
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Basic strategy
I Every computation is assigned a Godel number pe11 pe2
2 . . . pell .
I Each exponent ei is the Godel number of an equation.
I The first A equations e1,...,eA are the axioms.
I The rest eA+1,...,el are expected to be deductions fromprevious equations.
I We require functions which can detect such deductions.
DMACC John M. Gillespie 98
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Basic strategy
I Every computation is assigned a Godel number pe11 pe2
2 . . . pell .
I Each exponent ei is the Godel number of an equation.
I The first A equations e1,...,eA are the axioms.
I The rest eA+1,...,el are expected to be deductions fromprevious equations.
I We require functions which can detect such deductions.
DMACC John M. Gillespie 99
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Basic strategy
I We will construct a primitive recursive function τ1 such thatτ1(ω, i) = 0 if and only if ei = z(j , k) for j < i in thecomputation ω.
I
(∃j∃k)[ (j < i) ∧ (k < ω) ∧ (ei = z(j , k)) ]
I We will construct similar primitive recursive functions τ2 andτ3 for the other deduction steps n and sub.
I Then τ1(ω, i)τ2(ω, i)τ3(ω, i) will equal zero if and only if theith equation of ω is some proper deduction.
DMACC John M. Gillespie 100
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Basic strategy
I We will construct a primitive recursive function τ1 such thatτ1(ω, i) = 0 if and only if ei = z(j , k) for j < i in thecomputation ω.
I
(∃j∃k)[ (j < i) ∧ (k < ω) ∧ (ei = z(j , k)) ]
I We will construct similar primitive recursive functions τ2 andτ3 for the other deduction steps n and sub.
I Then τ1(ω, i)τ2(ω, i)τ3(ω, i) will equal zero if and only if theith equation of ω is some proper deduction.
DMACC John M. Gillespie 101
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Basic strategy
I We will construct a primitive recursive function τ1 such thatτ1(ω, i) = 0 if and only if ei = z(j , k) for j < i in thecomputation ω.
I
(∃j∃k)[ (j < i) ∧ (k < ω) ∧ (ei = z(j , k)) ]
I We will construct similar primitive recursive functions τ2 andτ3 for the other deduction steps n and sub.
I Then τ1(ω, i)τ2(ω, i)τ3(ω, i) will equal zero if and only if theith equation of ω is some proper deduction.
DMACC John M. Gillespie 102
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Basic strategy
I We will construct a primitive recursive function τ1 such thatτ1(ω, i) = 0 if and only if ei = z(j , k) for j < i in thecomputation ω.
I
(∃j∃k)[ (j < i) ∧ (k < ω) ∧ (ei = z(j , k)) ]
I We will construct similar primitive recursive functions τ2 andτ3 for the other deduction steps n and sub.
I Then τ1(ω, i)τ2(ω, i)τ3(ω, i) will equal zero if and only if theith equation of ω is some proper deduction.
DMACC John M. Gillespie 103
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Admissable deduction step z
Substituting zero for the variable xi in the equation e altarsthe Godel number as follows.
z(e, i) =
long(e)∏j=1
pj
1 if expj(e) = 2i + 9expj(e) else
This function is primitive recursive.
DMACC John M. Gillespie 104
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Admissable deduction step z
Substituting zero for the variable xi in the equation e altarsthe Godel number as follows.
z(e, i) =
long(e)∏j=1
pj
1 if expj(e) = 2i + 9expj(e) else
This function is primitive recursive.
DMACC John M. Gillespie 105
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Admissable deduction step z
Substituting zero for the variable xi in the equation e altarsthe Godel number as follows.
z(e, i) =
long(e)∏j=1
pj
1 if expj(e) = 2i + 9expj(e) else
This function is primitive recursive.
DMACC John M. Gillespie 106
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Admissable deduction step n
Substituting a non-zero value σ0(xi) in the variable xi ofequation e altars the Godel number as follows.
n(e, i) =
long(e)⊗j=1
{q if expj(e) = 2i + 9
pexpj(e)1 else
}
This function is primitive recursive. What is q?
DMACC John M. Gillespie 107
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Admissable deduction step n
Substituting a non-zero value σ0(xi) in the variable xi ofequation e altars the Godel number as follows.
n(e, i) =
long(e)⊗j=1
{q if expj(e) = 2i + 9
pexpj(e)1 else
}
This function is primitive recursive. What is q?
DMACC John M. Gillespie 108
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Admissable deduction step n
Substituting a non-zero value σ0(xi) in the variable xi ofequation e altars the Godel number as follows.
n(e, i) =
long(e)⊗j=1
{q if expj(e) = 2i + 9
pexpj(e)1 else
}
This function is primitive recursive.
What is q?
DMACC John M. Gillespie 109
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Admissable deduction step n
Substituting a non-zero value σ0(xi) in the variable xi ofequation e altars the Godel number as follows.
n(e, i) =
long(e)⊗j=1
{q if expj(e) = 2i + 9
pexpj(e)1 else
}
This function is primitive recursive. What is q?
DMACC John M. Gillespie 110
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Useful formulas
If m is the Godel number of an equation then eq(m) locates theequality sign.
eq(m) = µj[j ≤ long(m) ∧ expj(m) = 3]
Bounded search =⇒ PR.
eq(m) = 1 +
long(m)∑i=1
i∏j=1
sg(sg(expj(m) ·− 3) + sg(3 ·− expj(m))
)
DMACC John M. Gillespie 111
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Useful formulas
If m is the Godel number of an equation then eq(m) locates theequality sign.
eq(m) = µj[j ≤ long(m) ∧ expj(m) = 3]
Bounded search =⇒ PR.
eq(m) = 1 +
long(m)∑i=1
i∏j=1
sg(sg(expj(m) ·− 3) + sg(3 ·− expj(m))
)
DMACC John M. Gillespie 112
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Useful formulas
If m is the Godel number of an equation then β(m) denotes theGodel number of the left-hand side of the equation.
β(m) =
eq(m) ·−1∏j=1
pexpj (m)
j
DMACC John M. Gillespie 113
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Useful formulas
Then the right-hand side of equation m which we denote by γ(m)is the following.
γ(m) =
long(m) ·−eq(m)∏j=1
pexpj+eq(m)(m)
j
DMACC John M. Gillespie 114
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Does a given equation contain some term?
Here we are testing for a copy of β(m) in the expression nbeginning at location j + 1. Denote this predicate β(m, n, j).
long(β(m))∑i=1
sg( expi+j(n) ·− expi β(m) ) + sg( expi β(m) ·− expi+j(n) )
I Hence β(m, n, j) = 0 iff expi (β(m)) = expi+j(n) for every1 ≤ i ≤ long(β(m)).
I So β(m, n, j) = 0 iff the expression for n contains theexpression for β(m) precisely after the jth symbol.
I Note that β(m, n, j) 6= 0 if long(n)− j < long(β(m)).
DMACC John M. Gillespie 115
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Does a given equation contain some term?
Here we are testing for a copy of β(m) in the expression nbeginning at location j + 1. Denote this predicate β(m, n, j).
long(β(m))∑i=1
sg( expi+j(n) ·− expi β(m) ) + sg( expi β(m) ·− expi+j(n) )
I Hence β(m, n, j) = 0 iff expi (β(m)) = expi+j(n) for every1 ≤ i ≤ long(β(m)).
I So β(m, n, j) = 0 iff the expression for n contains theexpression for β(m) precisely after the jth symbol.
I Note that β(m, n, j) 6= 0 if long(n)− j < long(β(m)).
DMACC John M. Gillespie 116
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Does a given equation contain some term?
Here we are testing for a copy of β(m) in the expression nbeginning at location j + 1. Denote this predicate β(m, n, j).
long(β(m))∑i=1
sg( expi+j(n) ·− expi β(m) ) + sg( expi β(m) ·− expi+j(n) )
I Hence β(m, n, j) = 0 iff expi (β(m)) = expi+j(n) for every1 ≤ i ≤ long(β(m)).
I So β(m, n, j) = 0 iff the expression for n contains theexpression for β(m) precisely after the jth symbol.
I Note that β(m, n, j) 6= 0 if long(n)− j < long(β(m)).
DMACC John M. Gillespie 117
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Does a given equation contain some term?
Here we are testing for a copy of β(m) in the expression nbeginning at location j + 1. Denote this predicate β(m, n, j).
long(β(m))∑i=1
sg( expi+j(n) ·− expi β(m) ) + sg( expi β(m) ·− expi+j(n) )
I Hence β(m, n, j) = 0 iff expi (β(m)) = expi+j(n) for every1 ≤ i ≤ long(β(m)).
I So β(m, n, j) = 0 iff the expression for n contains theexpression for β(m) precisely after the jth symbol.
I Note that β(m, n, j) 6= 0 if long(n)− j < long(β(m)).
DMACC John M. Gillespie 118
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The binary function eq(m, n)
We will use eq also to denote the binary function given below.
eq(m, n) = sg(m ·− n) + sg(n ·−m)
Hence eq(m, n) = 0 if and only if m = n. Then
long(β(m))∑i=1
sg( expi+j(n) ·− expi β(m) ) + sg( expi β(m) ·− expi+j(n) )
simplifies to
β(m, n, j) =
long(β(m))∑i=1
eq(expi+j(n), expi β(m)).
DMACC John M. Gillespie 119
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The binary function eq(m, n)
We will use eq also to denote the binary function given below.
eq(m, n) = sg(m ·− n) + sg(n ·−m)
Hence eq(m, n) = 0 if and only if m = n.
Then
long(β(m))∑i=1
sg( expi+j(n) ·− expi β(m) ) + sg( expi β(m) ·− expi+j(n) )
simplifies to
β(m, n, j) =
long(β(m))∑i=1
eq(expi+j(n), expi β(m)).
DMACC John M. Gillespie 120
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The binary function eq(m, n)
We will use eq also to denote the binary function given below.
eq(m, n) = sg(m ·− n) + sg(n ·−m)
Hence eq(m, n) = 0 if and only if m = n. Then
long(β(m))∑i=1
sg( expi+j(n) ·− expi β(m) ) + sg( expi β(m) ·− expi+j(n) )
simplifies to
β(m, n, j) =
long(β(m))∑i=1
eq(expi+j(n), expi β(m)).
DMACC John M. Gillespie 121
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The binary function eq(m, n)
We will use eq also to denote the binary function given below.
eq(m, n) = sg(m ·− n) + sg(n ·−m)
Hence eq(m, n) = 0 if and only if m = n. Then
long(β(m))∑i=1
sg( expi+j(n) ·− expi β(m) ) + sg( expi β(m) ·− expi+j(n) )
simplifies to
β(m, n, j) =
long(β(m))∑i=1
eq(expi+j(n), expi β(m)).
DMACC John M. Gillespie 122
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Admissable deduction step sub
Substituting γ(m) for β(m) in equation n altars it’s Godelnumber as follows.
If β(m, n, j) = 0 then
sub(m, n, j) =
j∏i=1
pexpi (n)i ∗ γ(m) ∗
long(n) ·−(j+longβ(m))∏i=1
pexpj+longβ(m)+i(n)
i
otherwise sub(m, n, j) = n.
This is a primitive recursive function.
DMACC John M. Gillespie 123
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Admissable deduction step sub
Substituting γ(m) for β(m) in equation n altars it’s Godelnumber as follows.
If β(m, n, j) = 0 then
sub(m, n, j) =
j∏i=1
pexpi (n)i ∗ γ(m) ∗
long(n) ·−(j+longβ(m))∏i=1
pexpj+longβ(m)+i(n)
i
otherwise sub(m, n, j) = n.
This is a primitive recursive function.
DMACC John M. Gillespie 124
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Valid deduction
The natural number ω is the Godel number of a valid deduction iffor every i ≤ long(ω) one of the following is true;
I expi (ω) is the Godel number of one of the defining equations.
I There exists j < i and some k such thatexpi (ω) = z(expj(ω), k).
I There exists j < i and some k such thatexpi (ω) = n(expj(ω), k)).
I ∃m, n < i and j where expi (ω) = sub(expm(ω), expn(ω), j)).
DMACC John M. Gillespie 125
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Valid deduction
The natural number ω is the Godel number of a valid deduction iffor every i ≤ long(ω) one of the following is true;
I expi (ω) is the Godel number of one of the defining equations.
I There exists j < i and some k such thatexpi (ω) = z(expj(ω), k).
I There exists j < i and some k such thatexpi (ω) = n(expj(ω), k)).
I ∃m, n < i and j where expi (ω) = sub(expm(ω), expn(ω), j)).
DMACC John M. Gillespie 126
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Valid deduction
The natural number ω is the Godel number of a valid deduction iffor every i ≤ long(ω) one of the following is true;
I expi (ω) is the Godel number of one of the defining equations.
I There exists j < i and some k such thatexpi (ω) = z(expj(ω), k).
I There exists j < i and some k such thatexpi (ω) = n(expj(ω), k)).
I ∃m, n < i and j where expi (ω) = sub(expm(ω), expn(ω), j)).
DMACC John M. Gillespie 127
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Valid deduction
The natural number ω is the Godel number of a valid deduction iffor every i ≤ long(ω) one of the following is true;
I expi (ω) is the Godel number of one of the defining equations.
I There exists j < i and some k such thatexpi (ω) = z(expj(ω), k).
I There exists j < i and some k such thatexpi (ω) = n(expj(ω), k)).
I ∃m, n < i and j where expi (ω) = sub(expm(ω), expn(ω), j)).
DMACC John M. Gillespie 128
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Valid deduction
Let τ1(ω, i) denote the predicate
(∃j∃k)[ (j < i) ∧ (k ≤ ω) ∧ eq(expi ω, z(expj ω, k)) ]
I Thus τ1(ω, i) is primitive recursive and, as formulated below,equals zero iff the property holds true.
τ1(ω, i) =i ·−1∏j=1
ω∏k=1
eq( expi ω, z(expj ω, k) )
DMACC John M. Gillespie 129
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Valid deduction
Let τ1(ω, i) denote the predicate
(∃j∃k)[ (j < i) ∧ (k ≤ ω) ∧ eq(expi ω, z(expj ω, k)) ]
I Thus τ1(ω, i) is primitive recursive and, as formulated below,equals zero iff the property holds true.
τ1(ω, i) =i ·−1∏j=1
ω∏k=1
eq( expi ω, z(expj ω, k) )
DMACC John M. Gillespie 130
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Valid deduction
Let τ2(ω, i) denote the predicate
(∃j∃k)[ (j < i) ∧ (k ≤ ω) ∧ eq(expi ω,n(expj ω, k)) ]
I Thus τ2(ω, i) is primitive recursive and, as formulated below,equals zero iff the property holds true.
τ2(ω, i) =i ·−1∏j=1
ω∏k=1
eq( expi ω,n(expj ω, k) )
DMACC John M. Gillespie 131
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Valid deduction
Let τ2(ω, i) denote the predicate
(∃j∃k)[ (j < i) ∧ (k ≤ ω) ∧ eq(expi ω,n(expj ω, k)) ]
I Thus τ2(ω, i) is primitive recursive and, as formulated below,equals zero iff the property holds true.
τ2(ω, i) =i ·−1∏j=1
ω∏k=1
eq( expi ω,n(expj ω, k) )
DMACC John M. Gillespie 132
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Valid deduction
Let τ3(ω, i) denote the predicate
(∃m)(∃n)(∃j)[ (m, n < i) ∧ (j ≤ long(n)) ∧ expi ω = sub(expm ω, expn ω, j) ]
I Thus τ3(ω, i) is primitive recursive and, as formulated below,equals zero iff the property holds true.
τ3(ω, i) =i ·−1∏m=1
i ·−1∏n=1
long(n)∏j=1
eq(expi ω, sub(expm ω, expn ω, j))
DMACC John M. Gillespie 133
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Valid deduction
Let τ3(ω, i) denote the predicate
(∃m)(∃n)(∃j)[ (m, n < i) ∧ (j ≤ long(n)) ∧ expi ω = sub(expm ω, expn ω, j) ]
I Thus τ3(ω, i) is primitive recursive and, as formulated below,equals zero iff the property holds true.
τ3(ω, i) =i ·−1∏m=1
i ·−1∏n=1
long(n)∏j=1
eq(expi ω, sub(expm ω, expn ω, j))
DMACC John M. Gillespie 134
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Useful formula
The symbol sequence σ0(0) represents the natural number oneand in general if σn
0 (0) then σ0(σn0 (0)) represents n + 1.
The Godel numbers of these expressions have a primitiverecursive formula.
ζ(0) = p11 and ζ(n + 1) = p2
1p52 ∗ ζ(n) ∗ p7
1
DMACC John M. Gillespie 135
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Useful formula
The symbol sequence σ0(0) represents the natural number oneand in general if σn
0 (0) then σ0(σn0 (0)) represents n + 1.
The Godel numbers of these expressions have a primitiverecursive formula.
ζ(0) = p11 and ζ(n + 1) = p2
1p52 ∗ ζ(n) ∗ p7
1
DMACC John M. Gillespie 136
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Useful formula
The symbol sequence σ0(0) represents the natural number oneand in general if σn
0 (0) then σ0(σn0 (0)) represents n + 1.
The Godel numbers of these expressions have a primitiverecursive formula.
ζ(0) = p11 and ζ(n + 1) = p2
1p52 ∗ ζ(n) ∗ p7
1
DMACC John M. Gillespie 137
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The valid deduction of ϕ(a1, ..., ar)
Let τ4(a1, ..., ar , ω) denote the predicate
(∃a)[ (a < γ(explongω(ω)) ∧ eq(explongω(ω),Φ(a)) ]
I Where Φ(a) := p2s+91 p5
2 ∗ ζ(a1) ∗ . . . ∗ ζ(ar ) ∗ p71p
32 ∗ ζ(a))
τ4(a1, ..., ar , ω) =
γ(explongω(ω))∏a=0
eq(explongω(ω),Φ(a))
DMACC John M. Gillespie 138
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The valid deduction of ϕ(a1, ..., ar)
Let τ4(a1, ..., ar , ω) denote the predicate
(∃a)[ (a < γ(explongω(ω)) ∧ eq(explongω(ω),Φ(a)) ]
I Where Φ(a) := p2s+91 p5
2 ∗ ζ(a1) ∗ . . . ∗ ζ(ar ) ∗ p71p
32 ∗ ζ(a))
τ4(a1, ..., ar , ω) =
γ(explongω(ω))∏a=0
eq(explongω(ω),Φ(a))
DMACC John M. Gillespie 139
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PR detection of a valid computation
Let τ(a1, ..., ar , ω) denote the following formula
long(ω)∑i=1+A
τ1(ω, i)τ2(ω, i)τ3(ω, i) + τ4(a1, ..., ar , ω)
I Thus ω is the Godel number of a valid computation of ϕ(~a) iffτ(a1, ..., ar , ω) = 0.
I The function µω[τ(a1, ..., ar , ω) = 0] denotes the unboundedsearch for a valid computation of ϕ(~a).
DMACC John M. Gillespie 140
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PR detection of a valid computation
Let τ(a1, ..., ar , ω) denote the following formula
long(ω)∑i=1+A
τ1(ω, i)τ2(ω, i)τ3(ω, i) + τ4(a1, ..., ar , ω)
I Thus ω is the Godel number of a valid computation of ϕ(~a) iffτ(a1, ..., ar , ω) = 0.
I The function µω[τ(a1, ..., ar , ω) = 0] denotes the unboundedsearch for a valid computation of ϕ(~a).
DMACC John M. Gillespie 141
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PR detection of a valid computation
Let τ(a1, ..., ar , ω) denote the following formula
long(ω)∑i=1+A
τ1(ω, i)τ2(ω, i)τ3(ω, i) + τ4(a1, ..., ar , ω)
I Thus ω is the Godel number of a valid computation of ϕ(~a) iffτ(a1, ..., ar , ω) = 0.
I The function µω[τ(a1, ..., ar , ω) = 0] denotes the unboundedsearch for a valid computation of ϕ(~a).
DMACC John M. Gillespie 142
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Kleene normal form
Let ψ(ω) denote the function
µj [j < γ(explongω(ω)) ∧ ζ(j) = γ(explong(ω)(ω)) ]
Since j < γ(explongω(ω)) the function ψ is primitive recursive andfinally we have the normal form below.
ϕ = ψ(µω[τ(a1, ..., ar , ω) = 0])
Thus every GR function can be written as above where ψ is aPR function and τ is a PR predicate on which the Kleeneoperator acts.
DMACC John M. Gillespie 143