%------------------------------------------------------------------------------
% File : Prover9---2026-6A
% Problem : SYN412+1 : TPTP v9.3.1. Released v2.0.0.
% Transfm : none
% Format : tptp:raw
% Command : prover9 -casc 300 -f /export/starexec/sandbox/benchmark/theBenchmark.p
% Computer : n017.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 09:37:50 AM UTC 2026
% Result : Theorem 0.20s 0.47s
% Output : CNFRefutation 0.20s
% Verified :
% SZS Type : Refutation
% Derivation depth : 6
% Number of leaves : 1
% Syntax : Number of formulae : 9 ( 3 unt; 0 def)
% Number of atoms : 19 ( 0 equ)
% Maximal formula atoms : 4 ( 2 avg)
% Number of connectives : 22 ( 12 ~; 6 |; 2 &)
% ( 2 <=>; 0 =>; 0 <=; 0 <~>)
% Maximal formula depth : 7 ( 4 avg)
% Maximal term depth : 1 ( 1 avg)
% Number of predicates : 2 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 1 ( 1 usr; 1 con; 0-0 aty)
% Number of variables : 9 ( 2 sgn 4 !; 3 ?)
% Comments :
%------------------------------------------------------------------------------
fof(kalish255,conjecture,
~ ? [X] :
! [Y] :
( f(X,Y)
<=> ~ f(X,X) ),
file('theBenchmark.p',kalish255) ).
fof(kalish255_neg,negated_conjecture,
~ ~ ? [X] :
! [Y] :
( f(X,Y)
<=> ~ f(X,X) ),
inference(assume_negation,[status(cth)],[kalish255]) ).
fof(nnf_1,plain,
? [X] :
! [Y] :
( ( f(X,X)
| f(X,Y) )
& ( ~ f(X,X)
| ~ f(X,Y) ) ),
inference(fof_nnf,[status(thm)],[kalish255_neg]) ).
fof(sk_kalish255_neg,plain,
! [Y] :
( ( f(sK0,sK0)
| f(sK0,Y) )
& ( ~ f(sK0,sK0)
| ~ f(sK0,Y) ) ),
inference(skolemize,[status(esa),new_symbols(skolem,[sK0]),skolemize(X,sK0)],[nnf_1]) ).
cnf(c_2,negated_conjecture,
( ~ f(sK0,sK0)
| ~ f(sK0,A) ),
inference(split_conjunct,[status(thm)],[sk_kalish255_neg]) ).
cnf(c_3,negated_conjecture,
( f(sK0,sK0)
| f(sK0,A) ),
inference(split_conjunct,[status(thm)],[sk_kalish255_neg]) ).
cnf(c_4,plain,
~ f(sK0,sK0),
inference(factor,[status(thm)],[c_2]) ).
cnf(c_6,plain,
f(sK0,sK0),
inference(factor,[status(thm)],[c_3]) ).
cnf(c_5,plain,
$false,
inference(resolve,[status(thm)],[c_4,c_6]) ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.05 % Problem : SYN412+1 : TPTP v9.3.1. Released v2.0.0.
% 0.00/0.07 % Command : prover9 -casc 300 -f /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.20/0.44 % Computer : n017.cluster.edu
% 0.20/0.44 % Model : x86_64 x86_64
% 0.20/0.44 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.20/0.44 % Memory : 8046.5625MB
% 0.20/0.44 % OS : Linux 6.8.0-71-generic
% 0.20/0.44 % CPULimit : 300
% 0.20/0.44 % WCLimit : 300
% 0.20/0.44 % DateTime : Sat Sep 26 17:37:04 UTC 2026
% 0.20/0.45 % CPUTime :
% 0.20/0.45 % Prover9 (64) version 2026-6A, July 2026, CASC-J13.
% 0.20/0.45 % Process 1281573 was started by sandbox on n017,
% 0.20/0.45 % Sat Sep 26 17:37:04 2026
% 0.20/0.45 % The command was "/export/starexec/sandbox/solver/bin/prover9 -casc 300 -f /export/starexec/sandbox/benchmark/theBenchmark.p".
% 0.20/0.45
% 0.20/0.45 % From the command line: assign(max_seconds, 300).
% 0.20/0.47
% 0.20/0.47 % SZS status Theorem for theBenchmark
% 0.20/0.47
% 0.20/0.47 % Proof 1 at 0.00 (+ 0.00) seconds.
% 0.20/0.47 % Length of proof is 5.
% 0.20/0.47 % Level of proof is 3.
% 0.20/0.47 % Maximum clause weight is 6.000.
% 0.20/0.47 % Given clauses 0.
% 0.20/0.47
% 0.20/0.47 % SZS output start CNFRefutation for theBenchmark
% See solution above
%------------------------------------------------------------------------------