TPTP Problem File: CAT019-5.p
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- Solve Problem
%--------------------------------------------------------------------------
% File : CAT019-5 : TPTP v9.0.0. Released v1.0.0.
% Domain : Category Theory
% Problem : Axiom of Indiscernibles
% Version : [Sco79] axioms : Reduced & Augmented > Complete.
% English : [all z (x=z <-> y=z)] -> x=y.
% Refs : [Sco79] Scott (1979), Identity and Existence in Intuitionist L
% Source : [ANL]
% Names : p15.ver3.no1.in [ANL]
% Status : Satisfiable
% Rating : 0.56 v9.0.0, 0.50 v8.2.0, 0.70 v8.1.0, 0.62 v7.5.0, 0.67 v7.4.0, 0.64 v7.3.0, 0.67 v7.1.0, 0.62 v7.0.0, 0.57 v6.4.0, 0.29 v6.3.0, 0.25 v6.2.0, 0.30 v6.1.0, 0.44 v6.0.0, 0.43 v5.5.0, 0.50 v5.4.0, 0.80 v5.3.0, 0.78 v5.2.0, 0.80 v5.0.0, 0.78 v4.1.0, 0.71 v4.0.1, 0.80 v4.0.0, 0.50 v3.7.0, 0.33 v3.4.0, 0.50 v3.3.0, 0.33 v3.2.0, 0.80 v3.1.0, 0.67 v2.7.0, 0.33 v2.6.0, 0.86 v2.5.0, 0.50 v2.4.0, 0.67 v2.2.1, 0.75 v2.2.0, 0.67 v2.1.0, 1.00 v2.0.0
% Syntax : Number of clauses : 15 ( 5 unt; 0 nHn; 12 RR)
% Number of literals : 29 ( 12 equ; 15 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 3 ( 2 usr; 0 prp; 1-2 aty)
% Number of functors : 6 ( 6 usr; 3 con; 0-2 aty)
% Number of variables : 21 ( 2 sgn)
% SPC : CNF_SAT_RFO_EQU_NUE
% Comments : Axioms simplified by Art Quaife.
% : Assumes something exists.
%--------------------------------------------------------------------------
%----Include Scott's axioms for category theory
include('Axioms/CAT004-0.ax').
%--------------------------------------------------------------------------
cnf(assume_c_exists,hypothesis,
there_exists(c) ).
cnf(equality_of_a_and_b1,hypothesis,
( ~ there_exists(Z)
| a != Z
| b = Z ) ).
cnf(equality_of_a_and_b2,hypothesis,
( ~ there_exists(Z)
| a = Z
| b != Z ) ).
cnf(prove_a_equals_b,negated_conjecture,
a != b ).
%----These lemmas are used as demodulators by the ANL group
%input_clause(name,hypothesis,
% [++equal(domain(domain(X)),domain(X))]).
%input_clause(name,hypothesis,
% [++equal(codomain(domain(X)),domain(X))]).
%input_clause(name,hypothesis,
% [++equal(domain(codomain(X)),codomain(X))]).
%input_clause(name,hypothesis,
% [++equal(codomain(codomain(X)),codomain(X))]).
%input_clause(name,hypothesis,
% [++equal(domain(compose(X,Y)),domain(Y))]).
%input_clause(name,hypothesis,
% [++equal(codomain(compose(X,Y)),codomain(X))]).
%input_clause(name,hypothesis,
% [++equal(compose(domain(X),domain(X)),domain(X))]).
%input_clause(name,hypothesis,
% [++equal(compose(codomain(X),codomain(X)),codomain(X))]).
%--------------------------------------------------------------------------