TSTP Solution File: RNG102+2 by SInE---0.4
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- Process Solution
%------------------------------------------------------------------------------
% File : SInE---0.4
% Problem : RNG102+2 : TPTP v5.0.0. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : Source/sine.py -e eprover -t %d %s
% Computer : art02.cs.miami.edu
% Model : i686 i686
% CPU : Intel(R) Pentium(R) 4 CPU 2.80GHz @ 2793MHz
% Memory : 2018MB
% OS : Linux 2.6.26.8-57.fc8
% CPULimit : 300s
% DateTime : Sun Dec 26 02:22:48 EST 2010
% Result : Theorem 0.18s
% Output : CNFRefutation 0.18s
% Verified :
% SZS Type : Refutation
% Derivation depth : 8
% Number of leaves : 2
% Syntax : Number of formulae : 14 ( 6 unt; 0 def)
% Number of atoms : 37 ( 12 equ)
% Maximal formula atoms : 7 ( 2 avg)
% Number of connectives : 31 ( 8 ~; 3 |; 20 &)
% ( 0 <=>; 0 =>; 0 <=; 0 <~>)
% Maximal formula depth : 7 ( 3 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 4 ( 2 usr; 1 prp; 0-2 aty)
% Number of functors : 8 ( 8 usr; 6 con; 0-2 aty)
% Number of variables : 9 ( 0 sgn 2 !; 6 ?)
% Comments :
%------------------------------------------------------------------------------
fof(9,axiom,
( ? [X1] :
( aElement0(X1)
& sdtasdt0(xc,X1) = xx )
& aElementOf0(xx,slsdtgt0(xc))
& ? [X1] :
( aElement0(X1)
& sdtasdt0(xc,X1) = xy )
& aElementOf0(xy,slsdtgt0(xc))
& aElement0(xz) ),
file('/tmp/tmpmAKqYG/sel_RNG102+2.p_1',m__1933) ).
fof(40,conjecture,
? [X1] :
( aElement0(X1)
& sdtasdt0(xc,X1) = xy ),
file('/tmp/tmpmAKqYG/sel_RNG102+2.p_1',m__) ).
fof(42,negated_conjecture,
~ ? [X1] :
( aElement0(X1)
& sdtasdt0(xc,X1) = xy ),
inference(assume_negation,[status(cth)],[40]) ).
fof(90,plain,
( ? [X2] :
( aElement0(X2)
& sdtasdt0(xc,X2) = xx )
& aElementOf0(xx,slsdtgt0(xc))
& ? [X3] :
( aElement0(X3)
& sdtasdt0(xc,X3) = xy )
& aElementOf0(xy,slsdtgt0(xc))
& aElement0(xz) ),
inference(variable_rename,[status(thm)],[9]) ).
fof(91,plain,
( aElement0(esk6_0)
& sdtasdt0(xc,esk6_0) = xx
& aElementOf0(xx,slsdtgt0(xc))
& aElement0(esk7_0)
& sdtasdt0(xc,esk7_0) = xy
& aElementOf0(xy,slsdtgt0(xc))
& aElement0(xz) ),
inference(skolemize,[status(esa)],[90]) ).
cnf(94,plain,
sdtasdt0(xc,esk7_0) = xy,
inference(split_conjunct,[status(thm)],[91]) ).
cnf(95,plain,
aElement0(esk7_0),
inference(split_conjunct,[status(thm)],[91]) ).
fof(249,negated_conjecture,
! [X1] :
( ~ aElement0(X1)
| sdtasdt0(xc,X1) != xy ),
inference(fof_nnf,[status(thm)],[42]) ).
fof(250,negated_conjecture,
! [X2] :
( ~ aElement0(X2)
| sdtasdt0(xc,X2) != xy ),
inference(variable_rename,[status(thm)],[249]) ).
cnf(251,negated_conjecture,
( sdtasdt0(xc,X1) != xy
| ~ aElement0(X1) ),
inference(split_conjunct,[status(thm)],[250]) ).
cnf(280,plain,
~ aElement0(esk7_0),
inference(spm,[status(thm)],[251,94,theory(equality)]) ).
cnf(285,plain,
$false,
inference(rw,[status(thm)],[280,95,theory(equality)]) ).
cnf(286,plain,
$false,
inference(cn,[status(thm)],[285,theory(equality)]) ).
cnf(287,plain,
$false,
286,
[proof] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% % SZS status Started for /home/graph/tptp/TPTP/Problems/RNG/RNG102+2.p
% --creating new selector for []
% -running prover on /tmp/tmpmAKqYG/sel_RNG102+2.p_1 with time limit 29
% -prover status Theorem
% Problem RNG102+2.p solved in phase 0.
% % SZS status Theorem for /home/graph/tptp/TPTP/Problems/RNG/RNG102+2.p
% % SZS status Ended for /home/graph/tptp/TPTP/Problems/RNG/RNG102+2.p
% Solved 1 out of 1.
% # Problem is unsatisfiable (or provable), constructing proof object
% # SZS status Theorem
% # SZS output start CNFRefutation.
% See solution above
% # SZS output end CNFRefutation
%
%------------------------------------------------------------------------------