%------------------------------------------------------------------------------
% File     : cocATP---0.2.0
% Problem  : SEV187^5 : TPTP v6.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p

% Computer : n183.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 13:33:51 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  : SEV187^5 : TPTP v6.1.0. Released v4.0.0.
% % Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% % Computer : n183.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 08:24:46 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 0x1cdd758>, <kernel.Type object at 0x1cddc20>) of role type named a_type
% Using role type
% Declaring a:Type
% FOF formula (<kernel.Constant object at 0x1cdb1b8>, <kernel.Type object at 0x1cdd4d0>) of role type named b_type
% Using role type
% Declaring b:Type
% FOF formula (<kernel.Constant object at 0x1cdd5a8>, <kernel.Type object at 0x1cdd2d8>) of role type named c_type
% Using role type
% Declaring c:Type
% FOF formula (forall (S:((a->(b->(c->Prop)))->Prop)), ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True)) of role conjecture named cCS3_ALL_THM_pme
% Conjecture to prove = (forall (S:((a->(b->(c->Prop)))->Prop)), ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True)):Prop
% Parameter a_DUMMY:a.
% Parameter b_DUMMY:b.
% Parameter c_DUMMY:c.
% We need to prove ['(forall (S:((a->(b->(c->Prop)))->Prop)), ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True))']
% Parameter a:Type.
% Parameter b:Type.
% Parameter c:Type.
% Trying to prove (forall (S:((a->(b->(c->Prop)))->Prop)), ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True))
% Found I:True
% Found (fun (x:(forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True)))=> I) as proof of True
% Found (fun (S:((a->(b->(c->Prop)))->Prop)) (x:(forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True)))=> I) as proof of ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True)
% Found (fun (S:((a->(b->(c->Prop)))->Prop)) (x:(forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True)))=> I) as proof of (forall (S:((a->(b->(c->Prop)))->Prop)), ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True))
% Got proof (fun (S:((a->(b->(c->Prop)))->Prop)) (x:(forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True)))=> I)
% Time elapsed = 0.038902s
% node=2 cost=5.000000 depth=2
% ::::::::::::::::::::::
% % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p
% % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% (fun (S:((a->(b->(c->Prop)))->Prop)) (x:(forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True)))=> I)
% % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% EOF
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