%------------------------------------------------------------------------------
% File     : cocATP---0.2.0
% Problem  : SEV159^5 : TPTP v6.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p

% Computer : n090.star.cs.uiowa.edu
% Model    : x86_64 x86_64
% CPU      : Intel(R) Xeon(R) CPU E5-2609 0 2.40GHz
% Memory   : 32286.75MB
% OS       : Linux 2.6.32-431.20.3.el6.x86_64
% CPULimit : 300s
% DateTime : Thu Jul 17 13:33:49 EDT 2014

% Result   : Theorem 0.35s
% Output   : Proof 0.35s
% Verified : 
% SZS Type : None (Parsing solution fails)
% Syntax   : Number of formulae    : 0

% Comments : 
%------------------------------------------------------------------------------
%----ERROR: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% % Problem  : SEV159^5 : TPTP v6.1.0. Released v4.0.0.
% % Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% % Computer : n090.star.cs.uiowa.edu
% % Model    : x86_64 x86_64
% % CPU      : Intel(R) Xeon(R) CPU E5-2609 0 @ 2.40GHz
% % Memory   : 32286.75MB
% % OS       : Linux 2.6.32-431.20.3.el6.x86_64
% % CPULimit : 300
% % DateTime : Thu Jul 17 08:16:01 CDT 2014
% % CPUTime  : 0.35 
% Python 2.7.5
% Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox/benchmark/', '/export/starexec/sandbox/benchmark/']
% FOF formula (<kernel.Constant object at 0xd3d0e0>, <kernel.Type object at 0xd3d7e8>) of role type named a_type
% Using role type
% Declaring a:Type
% FOF formula (a->(forall (Xy:a), (((eq a) Xy) Xy))) of role conjecture named cTHM181_pme
% Conjecture to prove = (a->(forall (Xy:a), (((eq a) Xy) Xy))):Prop
% Parameter a_DUMMY:a.
% We need to prove ['(a->(forall (Xy:a), (((eq a) Xy) Xy)))']
% Parameter a:Type.
% Trying to prove (a->(forall (Xy:a), (((eq a) Xy) Xy)))
% Found eq_ref0:=(eq_ref a):(forall (a0:a), (((eq a) a0) a0))
% Found (eq_ref a) as proof of (forall (Xy:a), (((eq a) Xy) Xy))
% Found (fun (Xx:a)=> (eq_ref a)) as proof of (forall (Xy:a), (((eq a) Xy) Xy))
% Found (fun (Xx:a)=> (eq_ref a)) as proof of (a->(forall (Xy:a), (((eq a) Xy) Xy)))
% Got proof (fun (Xx:a)=> (eq_ref a))
% Time elapsed = 0.043552s
% node=3 cost=-94.000000 depth=2
% ::::::::::::::::::::::
% % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p
% % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% (fun (Xx:a)=> (eq_ref a))
% % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% EOF
%------------------------------------------------------------------------------