TPTP Problem File: ROB016-1.p
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%--------------------------------------------------------------------------
% File : ROB016-1 : TPTP v9.0.0. Released v1.0.0.
% Domain : Robbins Algebra
% Problem : If -(d + e) = -e then -(e + k(d + -(d + -e))) = -e, for k>0
% Version : [Win90] (equality) axioms.
% English :
% Refs : [Win90] Winker (1990), Robbins Algebra: Conditions that make a
% Source : [Win90]
% Names : Corollary 3.7 [Win90]
% Status : Unsatisfiable
% Rating : 0.00 v7.0.0, 0.14 v6.3.0, 0.17 v6.2.0, 0.00 v6.0.0, 0.11 v5.5.0, 0.50 v5.4.0, 0.47 v5.3.0, 0.67 v5.2.0, 0.12 v5.1.0, 0.14 v5.0.0, 0.29 v4.1.0, 0.11 v4.0.1, 0.00 v3.1.0, 0.11 v2.7.0, 0.00 v2.6.0, 0.29 v2.5.0, 0.00 v2.4.0, 0.00 v2.2.1, 0.22 v2.2.0, 0.14 v2.1.0, 0.20 v2.0.0
% Syntax : Number of clauses : 11 ( 8 unt; 0 nHn; 6 RR)
% Number of literals : 15 ( 9 equ; 5 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 8 ( 2 avg)
% Number of predicates : 2 ( 1 usr; 0 prp; 1-2 aty)
% Number of functors : 8 ( 8 usr; 4 con; 0-2 aty)
% Number of variables : 14 ( 0 sgn)
% SPC : CNF_UNS_RFO_SEQ_HRN
% Comments : The extra lemma is required for the proof.
%--------------------------------------------------------------------------
%----Include axioms for Robbins algebra
include('Axioms/ROB001-0.ax').
%----Include axioms for Robbins algebra numbers
include('Axioms/ROB001-1.ax').
%--------------------------------------------------------------------------
cnf(condition,hypothesis,
negate(add(d,e)) = negate(e) ).
cnf(k_positive,axiom,
positive_integer(k) ).
%----Lemma 3.6 required
cnf(lemma_3_6,axiom,
( negate(add(negate(Y),negate(add(X,negate(Y))))) != X
| ~ positive_integer(Vk)
| negate(add(Y,multiply(Vk,add(X,negate(add(X,negate(Y))))))) = negate(Y) ) ).
cnf(prove_result,negated_conjecture,
negate(add(e,multiply(k,add(d,negate(add(d,negate(e))))))) != negate(e) ).
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