TPTP Problem File: ALG246-1.p
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- Solve Problem
%------------------------------------------------------------------------------
% File : ALG246-1 : TPTP v9.0.0. Released v4.0.0.
% Domain : Group Theory (Quasigroups)
% Problem : Axioms of SBL algebras are not independent
% Version : Especial.
% English :
% Refs : [PS08] Phillips & Stanovsky (2008), Automated Theorem Proving
% : [Sta08] Stanovsky (2008), Email to G. Sutcliffe
% Source : [Sta08]
% Names : Sxx_1 [Sta08]
% Status : Unsatisfiable
% Rating : 0.73 v9.0.0, 0.68 v8.2.0, 0.71 v8.1.0, 0.80 v7.5.0, 0.79 v7.4.0, 0.87 v7.3.0, 0.79 v7.1.0, 0.78 v7.0.0, 0.74 v6.4.0, 0.79 v6.3.0, 0.82 v6.2.0, 0.86 v6.1.0, 0.88 v6.0.0, 0.90 v5.5.0, 0.84 v5.4.0, 0.80 v5.3.0, 0.75 v5.2.0, 0.79 v5.1.0, 0.80 v5.0.0, 0.79 v4.1.0, 0.73 v4.0.1, 0.79 v4.0.0
% Syntax : Number of clauses : 14 ( 14 unt; 0 nHn; 1 RR)
% Number of literals : 14 ( 14 equ; 1 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 5 ( 2 avg)
% Number of predicates : 1 ( 0 usr; 0 prp; 2-2 aty)
% Number of functors : 11 ( 11 usr; 4 con; 0-2 aty)
% Number of variables : 23 ( 1 sgn)
% SPC : CNF_UNS_RFO_PEQ_UEQ
% Comments : The LADR formulas contain function or predicate symbols that are
% not legal TPTP symbols, and we have replaced those symbols with
% new symbols. Here is the list of the bad symbols and the
% corresponding replacements.
% (arity 2) / -> tptp0
% (arity 2) * -> tptp1
% (arity 1) - -> tptp2
% (arity 0) 0 -> tptp3
% (arity 2) ^ -> tptp4
% (arity 0) 1 -> tptp5
%------------------------------------------------------------------------------
cnf(c01,axiom,
tptp1(A,B) = tptp1(B,A) ).
cnf(c02,axiom,
tptp1(tptp5,A) = A ).
cnf(c03,axiom,
tptp1(A,tptp0(A,B)) = tptp1(B,tptp0(B,A)) ).
cnf(c04,axiom,
tptp0(tptp1(A,B),C) = tptp0(A,tptp0(B,C)) ).
cnf(c05,axiom,
tptp0(tptp3,A) = tptp5 ).
cnf(c06,axiom,
tptp0(tptp0(tptp0(A,B),C),tptp0(tptp0(tptp0(B,A),C),C)) = tptp5 ).
cnf(c07,axiom,
tptp4(A,B) = tptp1(A,tptp0(A,B)) ).
cnf(c08,axiom,
v(A,B) = tptp4(tptp0(tptp0(A,B),B),tptp0(tptp0(B,A),A)) ).
cnf(c09,axiom,
tptp0(tptp1(A,B),tptp3) = v(tptp0(A,tptp3),tptp0(B,tptp3)) ).
cnf(c10,axiom,
m(A) = tptp0(A,tptp3) ).
cnf(c11,axiom,
n(A) = m(tptp2(A)) ).
cnf(c12,axiom,
tptp2(tptp2(A)) = A ).
cnf(c13,axiom,
n(tptp0(A,B)) = n(tptp0(tptp2(B),tptp2(A))) ).
cnf(goals,negated_conjecture,
v(tptp1(n(a),n(tptp0(a,b))),n(b)) != n(b) ).
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