TSTP Solution File: SYN045+1 by Etableau---0.67
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%------------------------------------------------------------------------------
% File : Etableau---0.67
% Problem : SYN045+1 : TPTP v8.1.0. Released v2.0.0.
% Transfm : none
% Format : tptp:raw
% Command : etableau --auto --tsmdo --quicksat=10000 --tableau=1 --tableau-saturation=1 -s -p --tableau-cores=8 --cpu-limit=%d %s
% Computer : n004.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit : 600s
% DateTime : Thu Jul 21 06:07:29 EDT 2022
% Result : Theorem 0.20s 0.38s
% Output : CNFRefutation 0.20s
% Verified :
% SZS Type : Refutation
% Derivation depth : 6
% Number of leaves : 1
% Syntax : Number of formulae : 14 ( 4 unt; 0 def)
% Number of atoms : 83 ( 0 equ)
% Maximal formula atoms : 44 ( 5 avg)
% Number of connectives : 108 ( 39 ~; 52 |; 15 &)
% ( 2 <=>; 0 =>; 0 <=; 0 <~>)
% Maximal formula depth : 15 ( 4 avg)
% Maximal term depth : 0 ( 0 avg)
% Number of predicates : 4 ( 3 usr; 4 prp; 0-0 aty)
% Number of functors : 0 ( 0 usr; 0 con; --- aty)
% Number of variables : 0 ( 0 sgn 0 !; 0 ?)
% Comments :
%------------------------------------------------------------------------------
fof(pel13,conjecture,
( ( p
| ( q
& r ) )
<=> ( ( p
| q )
& ( p
| r ) ) ),
file('/export/starexec/sandbox2/benchmark/theBenchmark.p',pel13) ).
fof(c_0_1,negated_conjecture,
~ ( ( p
| ( q
& r ) )
<=> ( ( p
| q )
& ( p
| r ) ) ),
inference(assume_negation,[status(cth)],[pel13]) ).
fof(c_0_2,negated_conjecture,
( ( ~ p
| ~ p
| ~ p )
& ( ~ r
| ~ p
| ~ p )
& ( ~ p
| ~ q
| ~ p )
& ( ~ r
| ~ q
| ~ p )
& ( ~ p
| ~ p
| ~ q
| ~ r )
& ( ~ r
| ~ p
| ~ q
| ~ r )
& ( ~ p
| ~ q
| ~ q
| ~ r )
& ( ~ r
| ~ q
| ~ q
| ~ r )
& ( p
| q
| q
| p )
& ( p
| r
| q
| p )
& ( p
| q
| r
| p )
& ( p
| r
| r
| p ) ),
inference(distribute,[status(thm)],[inference(fof_nnf,[status(thm)],[c_0_1])]) ).
cnf(c_0_3,negated_conjecture,
( p
| q
| q
| p ),
inference(split_conjunct,[status(thm)],[c_0_2]) ).
cnf(c_0_4,negated_conjecture,
( ~ p
| ~ p
| ~ p ),
inference(split_conjunct,[status(thm)],[c_0_2]) ).
cnf(c_0_5,negated_conjecture,
( p
| r
| r
| p ),
inference(split_conjunct,[status(thm)],[c_0_2]) ).
cnf(c_0_6,negated_conjecture,
( ~ r
| ~ q
| ~ q
| ~ r ),
inference(split_conjunct,[status(thm)],[c_0_2]) ).
cnf(c_0_7,negated_conjecture,
( p
| q ),
inference(cn,[status(thm)],[c_0_3]) ).
cnf(c_0_8,negated_conjecture,
~ p,
inference(cn,[status(thm)],[c_0_4]) ).
cnf(c_0_9,negated_conjecture,
( p
| r ),
inference(cn,[status(thm)],[c_0_5]) ).
cnf(c_0_10,negated_conjecture,
( ~ q
| ~ r ),
inference(cn,[status(thm)],[c_0_6]) ).
cnf(c_0_11,negated_conjecture,
q,
inference(sr,[status(thm)],[c_0_7,c_0_8]) ).
cnf(c_0_12,negated_conjecture,
r,
inference(sr,[status(thm)],[c_0_9,c_0_8]) ).
cnf(c_0_13,negated_conjecture,
$false,
inference(cn,[status(thm)],[inference(rw,[status(thm)],[inference(cn,[status(thm)],[inference(rw,[status(thm)],[c_0_10,c_0_11])]),c_0_12])]),
[proof] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.13 % Problem : SYN045+1 : TPTP v8.1.0. Released v2.0.0.
% 0.04/0.13 % Command : etableau --auto --tsmdo --quicksat=10000 --tableau=1 --tableau-saturation=1 -s -p --tableau-cores=8 --cpu-limit=%d %s
% 0.14/0.35 % Computer : n004.cluster.edu
% 0.14/0.35 % Model : x86_64 x86_64
% 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35 % Memory : 8042.1875MB
% 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35 % CPULimit : 300
% 0.14/0.35 % WCLimit : 600
% 0.14/0.35 % DateTime : Mon Jul 11 20:20:22 EDT 2022
% 0.14/0.35 % CPUTime :
% 0.20/0.38 # No SInE strategy applied
% 0.20/0.38 # Auto-Mode selected heuristic G_E___208_C18_F1_SE_CS_SP_PS_S5PRR_RG_S04AN
% 0.20/0.38 # and selection function SelectComplexExceptUniqMaxHorn.
% 0.20/0.38 #
% 0.20/0.38 # Presaturation interreduction done
% 0.20/0.38
% 0.20/0.38 # Proof found!
% 0.20/0.38 # SZS status Theorem
% 0.20/0.38 # SZS output start CNFRefutation
% See solution above
% 0.20/0.38 # Training examples: 0 positive, 0 negative
%------------------------------------------------------------------------------