%--------------------------------------------------------------------------
% File     : COL042-1 : TPTP v9.0.0. Released v1.0.0.
% Domain   : Combinatory Logic
% Problem  : Strong fixed point for B and W1
% Version  : [WM88] (equality) axioms.
% English  : The strong fixed point property holds for the set
%            P consisting of the combinators B and W1, where ((Bx)y)z
%            = x(yz), (W1x)y = (yx)x.

% Refs     : [Smu85] Smullyan (1978), To Mock a Mocking Bird and Other Logi
%          : [MW87]  McCune & Wos (1987), A Case Study in Automated Theorem
%          : [WM88]  Wos & McCune (1988), Challenge Problems Focusing on Eq
%          : [MW88]  McCune & Wos (1988), Some Fixed Point Problems in Comb
%          : [Wos93] Wos (1993), The Kernel Strategy and Its Use for the St
% Source   : [MW88]
% Names    : - [MW88]
%          : Question 5 [Wos93]

% Status   : Unsatisfiable
% Rating   : 0.82 v8.2.0, 0.79 v8.1.0, 0.90 v7.5.0, 0.88 v7.4.0, 0.87 v7.3.0, 0.89 v7.1.0, 0.83 v7.0.0, 0.84 v6.3.0, 0.82 v6.2.0, 0.86 v6.1.0, 0.88 v6.0.0, 0.95 v5.4.0, 0.87 v5.3.0, 0.83 v5.2.0, 0.79 v5.1.0, 0.80 v5.0.0, 0.79 v4.1.0, 0.73 v4.0.1, 0.71 v4.0.0, 0.69 v3.7.0, 0.67 v3.4.0, 0.75 v3.3.0, 0.86 v3.1.0, 0.89 v2.7.0, 0.91 v2.6.0, 0.83 v2.5.0, 0.75 v2.4.0, 0.33 v2.2.1, 0.89 v2.2.0, 1.00 v2.0.0
% Syntax   : Number of clauses     :    3 (   3 unt;   0 nHn;   1 RR)
%            Number of literals    :    3 (   3 equ;   1 neg)
%            Maximal clause size   :    1 (   1 avg)
%            Maximal term depth    :    4 (   2 avg)
%            Number of predicates  :    1 (   0 usr;   0 prp; 2-2 aty)
%            Number of functors    :    4 (   4 usr;   2 con; 0-2 aty)
%            Number of variables   :    6 (   0 sgn)
% SPC      : CNF_UNS_RFO_PEQ_UEQ

% Comments :
%--------------------------------------------------------------------------
cnf(b_definition,axiom,
    apply(apply(apply(b,X),Y),Z) = apply(X,apply(Y,Z)) ).

cnf(w1_definition,axiom,
    apply(apply(w1,X),Y) = apply(apply(Y,X),X) ).

cnf(prove_fixed_point,negated_conjecture,
    apply(Y,f(Y)) != apply(f(Y),apply(Y,f(Y))) ).

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