%--------------------------------------------------------------------------
% File     : PUZ023-1 : TPTP v8.2.0. 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 :
%--------------------------------------------------------------------------
%----Include axioms for truthtellers and liars
include('Axioms/PUZ002-0.ax').
%--------------------------------------------------------------------------
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) ).

%--------------------------------------------------------------------------