## TPTP Problem File: KLE104-10.p

View Solutions - Solve Problem

```%------------------------------------------------------------------------------
% File     : KLE104-10 : TPTP v7.5.0. Released v7.3.0.
% Domain   : Puzzles
% Problem  : Forward and backward boxes are conjugates
% 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   : 0.70 v7.5.0, 0.71 v7.4.0, 0.65 v7.3.0
% Syntax   : Number of clauses     :   31 (   0 non-Horn;  31 unit;   2 RR)
%            Number of atoms       :   31 (  31 equality)
%            Maximal clause size   :    1 (   1 average)
%            Number of predicates  :    1 (   0 propositional; 2-2 arity)
%            Number of functors    :   21 (   6 constant; 0-4 arity)
%            Number of variables   :   51 (   4 singleton)
%            Maximal term depth    :    6 (   2 average)
% SPC      : CNF_UNS_RFO_PEQ_UEQ

% Comments : Converted from KLE104+1 to UEQ using [CS18].
%------------------------------------------------------------------------------
cnf(ifeq_axiom,axiom,
( ifeq2(A,A,B,C) = B )).

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

( addition(A,zero) = A )).

( addition(A,A) = A )).

cnf(multiplicative_associativity,axiom,
( multiplication(A,multiplication(B,C)) = multiplication(multiplication(A,B),C) )).

cnf(multiplicative_right_identity,axiom,
( multiplication(A,one) = A )).

cnf(multiplicative_left_identity,axiom,
( multiplication(one,A) = A )).

cnf(right_distributivity,axiom,

cnf(left_distributivity,axiom,

cnf(right_annihilation,axiom,
( multiplication(A,zero) = zero )).

cnf(left_annihilation,axiom,
( multiplication(zero,A) = zero )).

cnf(order_1,axiom,
( ifeq(leq(A,B),true,addition(A,B),B) = B )).

cnf(order,axiom,
( ifeq2(addition(A,B),B,leq(A,B),true) = true )).

cnf(domain1,axiom,
( multiplication(antidomain(X0),X0) = zero )).

cnf(domain2,axiom,
( addition(antidomain(multiplication(X0,X1)),antidomain(multiplication(X0,antidomain(antidomain(X1))))) = antidomain(multiplication(X0,antidomain(antidomain(X1)))) )).

cnf(domain3,axiom,
( addition(antidomain(antidomain(X0)),antidomain(X0)) = one )).

cnf(domain4,axiom,
( domain(X0) = antidomain(antidomain(X0)) )).

cnf(codomain1,axiom,
( multiplication(X0,coantidomain(X0)) = zero )).

cnf(codomain2,axiom,
( addition(coantidomain(multiplication(X0,X1)),coantidomain(multiplication(coantidomain(coantidomain(X0)),X1))) = coantidomain(multiplication(coantidomain(coantidomain(X0)),X1)) )).

cnf(codomain3,axiom,
( addition(coantidomain(coantidomain(X0)),coantidomain(X0)) = one )).

cnf(codomain4,axiom,
( codomain(X0) = coantidomain(coantidomain(X0)) )).

cnf(complement,axiom,
( c(X0) = antidomain(domain(X0)) )).

cnf(domain_difference,axiom,
( domain_difference(X0,X1) = multiplication(domain(X0),antidomain(X1)) )).

cnf(forward_diamond,axiom,
( forward_diamond(X0,X1) = domain(multiplication(X0,domain(X1))) )).

cnf(backward_diamond,axiom,
( backward_diamond(X0,X1) = codomain(multiplication(codomain(X1),X0)) )).

cnf(forward_box,axiom,
( forward_box(X0,X1) = c(forward_diamond(X0,c(X1))) )).

cnf(backward_box,axiom,
( backward_box(X0,X1) = c(backward_diamond(X0,c(X1))) )).

cnf(goals,negated_conjecture,
( addition(domain(sK2_goals_X1),backward_box(sK3_goals_X0,domain(sK1_goals_X2))) = one )).

cnf(goals_1,negated_conjecture,
( addition(forward_box(sK3_goals_X0,domain(sK2_goals_X1)),domain(sK1_goals_X2)) != one )).

%------------------------------------------------------------------------------
```