%------------------------------------------------------------------------------
% File     : cocATP---0.2.0
% Problem  : SEU902^5 : TPTP v6.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p

% Computer : n117.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 13:33:21 EDT 2014

% Result   : Unknown 1.21s
% Output   : None 
% Verified : 
% SZS Type : None (Parsing solution fails)
% Syntax   : Number of formulae    : 0

% Comments : 
%------------------------------------------------------------------------------
%----NO SOLUTION OUTPUT BY SYSTEM
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% % Problem  : SEU902^5 : TPTP v6.1.0. Released v4.0.0.
% % Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% % Computer : n117.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 11:41:36 CDT 2014
% % CPUTime  : 1.21 
% 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 0x20e6320>, <kernel.DependentProduct object at 0x22c7518>) of role type named d
% Using role type
% Declaring d:(fofType->Prop)
% FOF formula (forall (Xh:((fofType->Prop)->fofType)), (((and (forall (Xp:(fofType->Prop)) (Xq:(fofType->Prop)), ((((eq fofType) (Xh Xp)) (Xh Xq))->(((eq (fofType->Prop)) Xp) Xq)))) (((eq (fofType->Prop)) d) (fun (Xz:fofType)=> ((ex (fofType->Prop)) (fun (Xt:(fofType->Prop))=> ((and ((Xt (Xh Xt))->False)) (((eq fofType) Xz) (Xh Xt))))))))->((d (Xh d))->False))) of role conjecture named cTHM143_pme
% Conjecture to prove = (forall (Xh:((fofType->Prop)->fofType)), (((and (forall (Xp:(fofType->Prop)) (Xq:(fofType->Prop)), ((((eq fofType) (Xh Xp)) (Xh Xq))->(((eq (fofType->Prop)) Xp) Xq)))) (((eq (fofType->Prop)) d) (fun (Xz:fofType)=> ((ex (fofType->Prop)) (fun (Xt:(fofType->Prop))=> ((and ((Xt (Xh Xt))->False)) (((eq fofType) Xz) (Xh Xt))))))))->((d (Xh d))->False))):Prop
% Parameter fofType_DUMMY:fofType.
% We need to prove ['(forall (Xh:((fofType->Prop)->fofType)), (((and (forall (Xp:(fofType->Prop)) (Xq:(fofType->Prop)), ((((eq fofType) (Xh Xp)) (Xh Xq))->(((eq (fofType->Prop)) Xp) Xq)))) (((eq (fofType->Prop)) d) (fun (Xz:fofType)=> ((ex (fofType->Prop)) (fun (Xt:(fofType->Prop))=> ((and ((Xt (Xh Xt))->False)) (((eq fofType) Xz) (Xh Xt))))))))->((d (Xh d))->False)))']
% Parameter fofType:Type.
% Parameter d:(fofType->Prop).
% Trying to prove (forall (Xh:((fofType->Prop)->fofType)), (((and (forall (Xp:(fofType->Prop)) (Xq:(fofType->Prop)), ((((eq fofType) (Xh Xp)) (Xh Xq))->(((eq (fofType->Prop)) Xp) Xq)))) (((eq (fofType->Prop)) d) (fun (Xz:fofType)=> ((ex (fofType->Prop)) (fun (Xt:(fofType->Prop))=> ((and ((Xt (Xh Xt))->False)) (((eq fofType) Xz) (Xh Xt))))))))->((d (Xh d))->False)))
% % SZS status GaveUp for /export/starexec/sandbox/benchmark/theBenchmark.p
% EOF
%------------------------------------------------------------------------------