%------------------------------------------------------------------------------
% File : Prover9---2026-6A
% Problem : KLE057+1 : TPTP v9.3.1. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : prover9 -casc 300 -f /export/starexec/sandbox/benchmark/theBenchmark.p
% Computer : n012.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8046.5625MB
% OS : Linux 6.8.0-71-generic
% CPULimit : 300s
% WCLimit : 300s
% DateTime : Sun Sep 27 07:44:13 AM UTC 2026
% Result : Theorem 0.04s 0.32s
% Output : CNFRefutation 0.04s
% Verified :
% SZS Type : Refutation
% Derivation depth : 6
% Number of leaves : 2
% Syntax : Number of formulae : 10 ( 6 unt; 0 def)
% Number of atoms : 14 ( 13 equ)
% Maximal formula atoms : 2 ( 1 avg)
% Number of connectives : 9 ( 5 ~; 0 |; 2 &)
% ( 0 <=>; 2 =>; 0 <=; 0 <~>)
% Maximal formula depth : 4 ( 2 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 3 ( 3 usr; 2 con; 0-1 aty)
% Number of variables : 3 ( 0 sgn 2 !; 1 ?)
% Comments :
%------------------------------------------------------------------------------
fof(goals,conjecture,
! [X0] :
( X0 = zero
=> domain(X0) = zero ),
file('theBenchmark.p',goals) ).
fof(goals_neg,negated_conjecture,
~ ! [X0] :
( X0 = zero
=> domain(X0) = zero ),
inference(assume_negation,[status(cth)],[goals]) ).
fof(nnf_17,plain,
? [X0] :
( X0 = zero
& domain(X0) != zero ),
inference(fof_nnf,[status(thm)],[goals_neg]) ).
fof(sk_goals_neg,plain,
( sK0 = zero
& domain(sK0) != zero ),
inference(skolemize,[status(esa),new_symbols(skolem,[sK0]),skolemize(X0,sK0)],[nnf_17]) ).
cnf(c_20,negated_conjecture,
sK0 = zero,
inference(split_conjunct,[status(thm)],[sk_goals_neg]) ).
cnf(c_21,negated_conjecture,
domain(sK0) != zero,
inference(split_conjunct,[status(thm)],[sk_goals_neg]) ).
cnf(c_22,negated_conjecture,
domain(zero) != zero,
inference(paramod,[status(thm)],[c_20,c_21]) ).
fof(domain4,axiom,
domain(zero) = zero,
file('theBenchmark.p',domain4) ).
cnf(c_24,plain,
domain(zero) = zero,
inference(clausify,[status(thm)],[domain4]) ).
cnf(c_25,plain,
$false,
inference(resolve,[status(thm)],[c_24,c_22]) ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.01 % Problem : KLE057+1 : TPTP v9.3.1. Released v4.0.0.
% 0.00/0.02 % Command : prover9 -casc 300 -f /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.04/0.31 % Computer : n012.cluster.edu
% 0.04/0.31 % Model : x86_64 x86_64
% 0.04/0.31 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.04/0.31 % Memory : 8046.5625MB
% 0.04/0.31 % OS : Linux 6.8.0-71-generic
% 0.04/0.31 % CPULimit : 300
% 0.04/0.31 % WCLimit : 300
% 0.04/0.31 % DateTime : Fri Sep 25 19:18:50 UTC 2026
% 0.04/0.31 % CPUTime :
% 0.04/0.31 % Prover9 (64) version 2026-6A, July 2026, CASC-J13.
% 0.04/0.31 % Process 2756577 was started by sandbox on n012,
% 0.04/0.31 % Fri Sep 25 19:18:50 2026
% 0.04/0.31 % The command was "/export/starexec/sandbox/solver/bin/prover9 -casc 300 -f /export/starexec/sandbox/benchmark/theBenchmark.p".
% 0.04/0.31
% 0.04/0.31 % From the command line: assign(max_seconds, 300).
% 0.04/0.32
% 0.04/0.32 % SZS status Theorem for theBenchmark
% 0.04/0.32
% 0.04/0.32 % Proof 1 at 0.00 (+ 0.00) seconds.
% 0.04/0.32 % Length of proof is 6.
% 0.04/0.32 % Level of proof is 3.
% 0.04/0.32 % Maximum clause weight is 4.000.
% 0.04/0.32 % Given clauses 0.
% 0.04/0.32
% 0.04/0.32 % SZS output start CNFRefutation for theBenchmark
% See solution above
%------------------------------------------------------------------------------