TPTP Problem File: ROB015-10.p

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```%------------------------------------------------------------------------------
% File     : ROB015-10 : TPTP v7.5.0. Released v7.5.0.
% Domain   : Puzzles
% Problem  : If -(-e + -(d + -e)) = d then -(e + k(d + -(d + -e))) = -e
% Version  : Especial.
% English  :

% Refs     : [CS18]  Claessen & Smallbone (2018), Efficient Encodings of Fi
%          : [Sma18] Smallbone (2018), Email to Geoff Sutcliffe
% Source   : [Sma18]
% Names    :

% Status   : Unsatisfiable
% Rating   : 0.65 v7.5.0
% Syntax   : Number of clauses     :   18 (   0 non-Horn;  18 unit;   6 RR)
%            Number of atoms       :   18 (  18 equality)
%            Maximal clause size   :    1 (   1 average)
%            Number of predicates  :    1 (   0 propositional; 2-2 arity)
%            Number of functors    :   16 (   7 constant; 0-4 arity)
%            Number of variables   :   30 (   4 singleton)
%            Maximal term depth    :    9 (   3 average)
% SPC      : CNF_UNS_RFO_PEQ_UEQ

% Comments : Converted from ROB015-2 to UEQ using [CS18].
%------------------------------------------------------------------------------
cnf(ifeq_axiom,axiom,
( ifeq4(A,A,B,C) = B )).

cnf(ifeq_axiom_001,axiom,
( ifeq3(A,A,B,C) = B )).

cnf(ifeq_axiom_002,axiom,
( ifeq2(A,A,B,C) = B )).

cnf(ifeq_axiom_003,axiom,
( ifeq(A,A,B,C) = B )).

cnf(robbins_axiom,axiom,

cnf(one_times_x,axiom,
( multiply(one,X) = X )).

cnf(one,axiom,
( positive_integer(one) = true )).

cnf(next_integer,axiom,
( ifeq(positive_integer(X),true,positive_integer(successor(X)),true) = true )).

cnf(lemma_3_2,axiom,

cnf(lemma_3_4,axiom,

cnf(condition,hypothesis,

cnf(k_positive,axiom,
( positive_integer(k) = true )).

cnf(base_step,axiom,