TPTP Problem File: PUZ023-1.p
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%--------------------------------------------------------------------------
% File : PUZ023-1 : TPTP v9.0.0. Released v1.0.0.
% Domain : Puzzles
% Problem : Knights and Knaves #27
% Version : Especial.
% English : There is an island with exactly two types of people :
% truthtellers who always tell the truth and liars who always
% lie. There are a group of three people, A, B, and C on the
% island. A stranger passes by and asks A, "How many
% truthtellers are among you ?" A answers indistinctly. So the
% stranger asks B, "what did A say?". B replies "A said that
% there is exactly one truthteller among us." Then C says,
% "Don't believe B; he is lying!" What are B and C. Answer:
% B is a liar and C is a truth-teller.
% Refs : [Smu78] Smullyan (1978), What is the Name of this Book?
% Source : [ANL]
% Names : Problem 27 [Smu78]
% : tandl27.ver1.in [ANL]
% Status : Unsatisfiable
% Rating : 0.00 v2.0.0
% Syntax : Number of clauses : 22 ( 5 unt; 6 nHn; 21 RR)
% Number of literals : 65 ( 0 equ; 39 neg)
% Maximal clause size : 5 ( 2 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 3 ( 3 usr; 0 prp; 1-3 aty)
% Number of functors : 12 ( 12 usr; 9 con; 0-2 aty)
% Number of variables : 32 ( 4 sgn)
% SPC : CNF_UNS_RFO_NEQ_NHN
% Comments :
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%----Include axioms for truthtellers and liars
include('Axioms/PUZ002-0.ax').
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cnf(one_is_the_truthteller,axiom,
( ~ people(X,Y,Z)
| ~ a_truth(one_truthteller)
| a_truth(truthteller(X))
| a_truth(truthteller(Y))
| a_truth(truthteller(Z)) ) ).
cnf(two_truthtellers1,axiom,
( ~ people(X,Y,Z)
| ~ a_truth(truthteller(X))
| ~ a_truth(truthteller(Y))
| ~ a_truth(one_truthteller) ) ).
cnf(two_truthtellers2,axiom,
( ~ people(X,Y,Z)
| ~ a_truth(truthteller(X))
| ~ a_truth(truthteller(Z))
| ~ a_truth(one_truthteller) ) ).
cnf(two_truthtellers3,axiom,
( ~ people(X,Y,Z)
| ~ a_truth(truthteller(Y))
| ~ a_truth(truthteller(Z))
| ~ a_truth(one_truthteller) ) ).
cnf(identify_one_truthteller1,axiom,
( ~ people(X,Y,Z)
| a_truth(one_truthteller)
| ~ a_truth(truthteller(X))
| a_truth(truthteller(Y))
| a_truth(truthteller(Z)) ) ).
cnf(identify_one_truthteller2,axiom,
( ~ people(X,Y,Z)
| a_truth(one_truthteller)
| ~ a_truth(truthteller(Y))
| a_truth(truthteller(X))
| a_truth(truthteller(Z)) ) ).
cnf(identify_one_truthteller3,axiom,
( ~ people(X,Y,Z)
| a_truth(one_truthteller)
| ~ a_truth(truthteller(Z))
| a_truth(truthteller(Y))
| a_truth(truthteller(X)) ) ).
cnf(a_b_and_c_are_people,hypothesis,
people(a,b,c) ).
cnf(a_says_garbage,hypothesis,
a_truth(says(a,garbage)) ).
cnf(b_says_a_says_one_truthteller,hypothesis,
a_truth(says(b,says(a,one_truthteller))) ).
cnf(c_says_b_lies,hypothesis,
a_truth(says(c,liar(b))) ).
%----This is an honest way of doing this. A simpler version could simply
%----prove that B is a liar and C is a truth-teller.
cnf(b_and_c_liars,hypothesis,
( ~ a_truth(liar(b))
| ~ a_truth(liar(c))
| an_answer(b_and_c_liars) ) ).
cnf(b_liar_and_c_truthteller,hypothesis,
( ~ a_truth(liar(b))
| ~ a_truth(truthteller(c))
| an_answer(b_liar_and_c_truthteller) ) ).
cnf(b_truthteller_and_c_liar,hypothesis,
( ~ a_truth(truthteller(b))
| ~ a_truth(liar(c))
| an_answer(b_truthteller_and_c_liar) ) ).
cnf(b_and_c_truthtellers,hypothesis,
( ~ a_truth(truthteller(b))
| ~ a_truth(truthteller(c))
| an_answer(b_and_c_truthtellers) ) ).
cnf(prove_there_is_an_answer,negated_conjecture,
~ an_answer(X) ).
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