TPTP Problem File: KLE110-10.p
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- Solve Problem
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% File : KLE110-10 : TPTP v9.0.0. Released v7.3.0.
% Domain : Puzzles
% Problem : Forward diamonds and backward boxes satisfy a cancellation law
% 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 : 0.86 v8.2.0, 0.88 v8.1.0, 0.95 v7.5.0, 0.96 v7.3.0
% Syntax : Number of clauses : 30 ( 30 unt; 0 nHn; 1 RR)
% Number of literals : 30 ( 30 equ; 1 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 6 ( 2 avg)
% Number of predicates : 1 ( 0 usr; 0 prp; 2-2 aty)
% Number of functors : 20 ( 20 usr; 5 con; 0-4 aty)
% Number of variables : 51 ( 4 sgn)
% SPC : CNF_UNS_RFO_PEQ_UEQ
% Comments : Converted from KLE110+1 to UEQ using [CS18].
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cnf(ifeq_axiom,axiom,
ifeq2(A,A,B,C) = B ).
cnf(ifeq_axiom_001,axiom,
ifeq(A,A,B,C) = B ).
cnf(additive_commutativity,axiom,
addition(A,B) = addition(B,A) ).
cnf(additive_associativity,axiom,
addition(A,addition(B,C)) = addition(addition(A,B),C) ).
cnf(additive_identity,axiom,
addition(A,zero) = A ).
cnf(additive_idempotence,axiom,
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,
multiplication(A,addition(B,C)) = addition(multiplication(A,B),multiplication(A,C)) ).
cnf(left_distributivity,axiom,
multiplication(addition(A,B),C) = addition(multiplication(A,C),multiplication(B,C)) ).
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_X0),backward_box(sK1_goals_X1,forward_diamond(sK1_goals_X1,domain(sK2_goals_X0)))) != backward_box(sK1_goals_X1,forward_diamond(sK1_goals_X1,domain(sK2_goals_X0))) ).
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