otter 2014-12-15-01-slideshare

OTTER(Organized Techniques for Theorem-proving and

Effective Research)

Heuristics computing 2014/12/15 9:25 – 11:00

Keio University, SFC

Four interactions and atom

定数は項である 変数は項である

f が n-変数関数記号， t1，･･･， tn が項であれば， f ( t1，･･･， tn ) は項である . term

P を n-変数述語記号， t1，･･･， tn を項とすると，P ( t1，･･･， tn ) は原子式である．また，原子式あるいは原子式の否定をリテラル という．

atom

nirmukta.com

Four interactions and OTTER

+

+ +-

-

+Gravitational interaction(weighting)weight_list(pick_and_purge).weight(P(rew($(0),$(0)),$(1)), 99)

Electromagnetic interaction (resolution)-father(x,y) | -mother(x,z) | grandmother(x,z)

Strong / weak interaction (modulation)(a=b) | (c=d).P(a) | Q(n, f(c,g(d))).

+

+

15 puzzleThe 15-puzzle (also called Gem Puzzle, Boss Puzzle, Game of Fifteen, Mystic Square and many others) is a sliding puzzle that consists of a frame of numbered square tiles in random order with one tile missing.

The n-puzzle is a classical problem for modelling algorithms involving heuristics.

http://mathworld.wolfram.com/15Puzzle.html

set(para_into).

list(usable).EQUAL(l(hole,l(n(x),y)),l(n(x),l(hole,y))).EQUAL(l(hole,l(x,l(y,l(z,l(u,l(n(w),v)))))),l(n(w),l(x,l(y,l(z,l(u,l(hole,v))))))).

-STATE(l(n(1),l(n(2),l(n(3),l(n(4),l(end,l(n(5),l(n(6),l(n(7),l(n(8),l(end,l(n(9),l(n(10),l(n(11),l(n(12),l(end,l(n(13),l(n(14),l(n(15),l(hole,end)))))))))))))))))))).end_of_list.

list(sos).STATE(l(n(1),l(n(6),l(n(2),l(n(4),l(end,l(n(5),l(hole,l(n(3),l(n(8),l(end,l(n(9),l(n(10),l(n(7),l(n(11),l(end,l(n(13),l(n(14),l(n(15),l(n(12),end)))))))))))))))))))).end_of_list.

Play the Fifteen puzzle onlinehttp://migo.sixbit.org/puzzles/fifteen/

paramodulation

paramodulation

1 |set(para_into).4 |assign(stats_level,1).5 |6 |list(usable).7 | a1 = b1.8 | a2 = b2.9 |end_of_list.10 |11 |list(sos).12 | P(a1,a2).13 |end_of_list.

Clauses are generated according to line 7 and 8.

Paramodulation

1 |set(para_into).2 |assign(max_given,3).3 |assign(stats_level,1).4 |5 |list(usable).6 | (a=b) | (c=d).7 |end_of_list.8 |9 |list(sos).10 | P(a) | Q(n, f(c,g(d))).11 |end_of_list.

1 |set(para_into).2 |assign(stats_level,1).3 |4 |list(usable).5 | father(ken)=jim.6 | mother(jim)=fay.7 |end_of_list.8 |9 |list(sos).10 | Grandmother (mother(father(x)),x).11 |end_of_list.

Paramodulation and demodulation

set(para_into).assign(stats_level,1).list(usable). father(ken)=jim. mother(jim)=fay.end_of_list.

list(sos). Grandmother(mother(father(x)),x).end_of_list.list(demodulators). mother(jim)=fay.end_of_list.

Liar paradox: 正直者と嘘つきの島

the liar paradox or liar's paradox (pseudomenon in Ancient Greek) is the statement "this sentence is false." Trying to assign to this statement a classical binary truth value leads to a contradiction. If "this sentence is false" is true, then the sentence is false, which is a contradiction. Conversely, if "this sentence is false" is false, then the sentence is true, which is also a contradiction.

%assign(max_proofs,-1).set(hyper_res).set(print_lists_at_end).assign(stats_level,0).

list(usable).-P(T(x)) | -P(Says(x,y)) | P(y).-P(L(x)) | -P(Says(x,y)) | -P(y).P(T(x)) | P(L(x)).-P(T(x)) | -P(L(x)).end_of_list.

list(sos).P(Says(A, L(A))).end_of_list.

Truth teller, Liar and Spy (Normal)

Problem formulation: N (power set) = P Q∨

P Q list(usable). -P(T(x)) | -P(Says(x,y)) | P(y). -P(L(x)) | -P(Says(x,y)) | Q(y). P(T(x)) | P(L(x)) | P(N(x)). -P(T(x)) | -P(L(x)). -P(T(x)) | -P(N(x)). -P(L(x)) | -P(N(x)). -P(n(T(x))) | P(L(x)) | P(N(x)). -P(n(L(x))) | P(T(x)) | P(N(x)). -Q(n(y)) | P(y). end_of_list. list(sos). P(Says(A,n(L(B)))). P(Says(B,n(T(A)))). end_of_list.

N

-Q(n(y)) | P(y). :　排中律Law of excluded middle

NP

Q

Law of excluded middle 排中律

Hilbert’s paradox of the Grand Hotel

Hilbert's paradox of the Grand Hotel is a veridical paradox (a valid argument with a seemingly absurd conclusion, as opposed to a falsidical paradox, which is a seemingly valid demonstration of an actual contradiction) about infinite sets meant to illustrate certain counterintuitive properties of infinite sets. Consider a hypothetical hotel with a countably infinite number of rooms, all of which are occupied. One might be tempted to think that the hotel would not be able to accommodate any newly arriving guests, as would be the case with a finite number of rooms.

http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel

Clause representation

IF A THEN B : -A | B

IF A & B THEN B : -A | -B | C

IF A THEN B or C : -A | B | C

achievable(state(hls(12),hs(0),ls(0),s(0),rem(3))).

-solvable(state(hls(12),hs(0),ls(0),s(0),rem(3))).

a b

a b c

a b

c

Proof and Satisfiability

Always true Always false

satisfiable

unsatisfiable

unsatisfiablesatisfiable

AB

C

A 　∩　 B 　→　 empty setB∪ ￢ A 　→ not Ф

Syntax “-A | B”: proof by contradiction

AB

C

A　⊃　 BB → A

There exists Ai where B∪ ￢ A → Ф

1 [] -Greek(x)|Person(x).2 [] -Person(x)|Mortal(x).3 [] Greek(socrates).4 [] -Mortal(socrates).5 [hyper,3,1] Person(socrates).6 [hyper,5,2] Mortal(socrates).7 [binary,6.1,4.1] $F.

3 |list(usable). 4 | -Greek(x) | Person(x). 5 | -Person(x) | Mortal(x). 6 |end_of_list. 8 |list(sos). 9 | Greek(socrates). 10 |end_of_list. 12 |list(passive). 13 | -Mortal(socrates). 14 |end_of_list.