TPTP Problem File: GRP002-2.p
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%--------------------------------------------------------------------------
% File : GRP002-2 : TPTP v9.0.0. Bugfixed v1.2.1.
% Domain : Group Theory
% Problem : Commutator equals identity in groups of order 3
% Version : [MOW76] (equality) axioms.
% English : In a group, if (for all x) the cube of x is the identity
% (i.e. a group of order 3), then the equation [[x,y],y]=
% identity holds, where [x,y] is the product of x, y, the
% inverse of x and the inverse of y (i.e. the commutator
% of x and y).
% Refs : [MOW76] McCharen et al. (1976), Problems and Experiments for a
% Source : [ANL]
% Names : commutator.ver2.in [ANL]
% Status : Unsatisfiable
% Rating : 0.05 v8.2.0, 0.12 v8.1.0, 0.15 v7.5.0, 0.12 v7.4.0, 0.22 v7.3.0, 0.11 v7.1.0, 0.00 v7.0.0, 0.05 v6.4.0, 0.11 v6.3.0, 0.12 v6.2.0, 0.14 v6.1.0, 0.06 v6.0.0, 0.29 v5.5.0, 0.21 v5.4.0, 0.00 v5.2.0, 0.07 v5.1.0, 0.13 v5.0.0, 0.14 v4.1.0, 0.18 v4.0.1, 0.14 v4.0.0, 0.15 v3.7.0, 0.11 v3.4.0, 0.12 v3.3.0, 0.00 v2.2.1, 0.22 v2.2.0, 0.29 v2.1.0, 0.29 v2.0.0
% Syntax : Number of clauses : 12 ( 12 unt; 0 nHn; 6 RR)
% Number of literals : 12 ( 12 equ; 1 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 3 ( 2 avg)
% Number of predicates : 1 ( 0 usr; 0 prp; 2-2 aty)
% Number of functors : 10 ( 10 usr; 8 con; 0-2 aty)
% Number of variables : 8 ( 0 sgn)
% SPC : CNF_UNS_RFO_PEQ_UEQ
% Comments :
% Bugfixes : v1.2.1 - Clause x_cubed_is_identity fixed.
%--------------------------------------------------------------------------
include('Axioms/GRP004-0.ax').
%--------------------------------------------------------------------------
%----Redundant two axioms, but established in standard axiomatizations.
cnf(right_identity,axiom,
multiply(X,identity) = X ).
cnf(right_inverse,axiom,
multiply(X,inverse(X)) = identity ).
%----This hypothesis is omitted in the ANL source version
cnf(x_cubed_is_identity,hypothesis,
multiply(X,multiply(X,X)) = identity ).
cnf(a_times_b_is_c,negated_conjecture,
multiply(a,b) = c ).
cnf(c_times_inverse_a_is_d,negated_conjecture,
multiply(c,inverse(a)) = d ).
cnf(d_times_inverse_b_is_h,negated_conjecture,
multiply(d,inverse(b)) = h ).
cnf(h_times_b_is_j,negated_conjecture,
multiply(h,b) = j ).
cnf(j_times_inverse_h_is_k,negated_conjecture,
multiply(j,inverse(h)) = k ).
cnf(prove_k_times_inverse_b_is_e,negated_conjecture,
multiply(k,inverse(b)) != identity ).
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