TPTP Problem File: BOO025-1.p
View Solutions
- Solve Problem
%--------------------------------------------------------------------------
% File : BOO025-1 : TPTP v9.0.0. Released v2.2.0.
% Domain : Boolean Algebra
% Problem : Half of Padmanabhan's 6-basis with Pixley, part 3.
% Version : [MP96] (equality) axioms : Especial.
% English : Part 3 (of 3) of the proof that half of Padmanaban's self-dual
% independent 6-basis for Boolean Algebra, together with a Pixley
% polynomial, is a basis for Boolean algebra.
% Refs : [McC98] McCune (1998), Email to G. Sutcliffe
% : [MP96] McCune & Padmanabhan (1996), Automated Deduction in Eq
% Source : [McC98]
% Names : DUAL-BA-2-c [MP96]
% Status : Unsatisfiable
% Rating : 0.14 v8.2.0, 0.17 v8.1.0, 0.20 v7.5.0, 0.21 v7.4.0, 0.22 v7.3.0, 0.21 v7.1.0, 0.11 v7.0.0, 0.16 v6.4.0, 0.21 v6.3.0, 0.24 v6.2.0, 0.29 v6.1.0, 0.31 v6.0.0, 0.43 v5.5.0, 0.42 v5.4.0, 0.27 v5.3.0, 0.17 v5.2.0, 0.21 v5.1.0, 0.27 v5.0.0, 0.21 v4.1.0, 0.18 v4.0.1, 0.21 v4.0.0, 0.23 v3.7.0, 0.11 v3.4.0, 0.12 v3.3.0, 0.14 v3.2.0, 0.07 v3.1.0, 0.11 v2.7.0, 0.00 v2.2.1
% Syntax : Number of clauses : 8 ( 8 unt; 0 nHn; 1 RR)
% Number of literals : 8 ( 8 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 : 7 ( 7 usr; 3 con; 0-3 aty)
% Number of variables : 15 ( 2 sgn)
% SPC : CNF_UNS_RFO_PEQ_UEQ
% Comments :
%--------------------------------------------------------------------------
%----Half of Padmanabhan's self-dual independent 6-basis for Boolean Algebra:
cnf(multiply_add,axiom,
multiply(add(X,Y),Y) = Y ).
cnf(multiply_add_property,axiom,
multiply(X,add(Y,Z)) = add(multiply(Y,X),multiply(Z,X)) ).
cnf(additive_inverse,axiom,
add(X,inverse(X)) = n1 ).
%----pixley(X,Y,Z) is a Pixley polynomial:
cnf(pixley_defn,axiom,
pixley(X,Y,Z) = add(multiply(X,inverse(Y)),add(multiply(X,Z),multiply(inverse(Y),Z))) ).
cnf(pixley1,axiom,
pixley(X,X,Y) = Y ).
cnf(pixley2,axiom,
pixley(X,Y,Y) = X ).
cnf(pixley3,axiom,
pixley(X,Y,X) = X ).
%----Denial of conclusion:
cnf(prove_equal_identity,negated_conjecture,
multiply(b,inverse(b)) != multiply(a,inverse(a)) ).
%--------------------------------------------------------------------------