TSTP Solution File: SEU562^2 by cocATP---0.2.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : cocATP---0.2.0
% Problem  : SEU562^2 : TPTP v6.1.0. Released v3.7.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p

% Computer : n111.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 13:32:28 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  : SEU562^2 : TPTP v6.1.0. Released v3.7.0.
% % Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% % Computer : n111.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 10:44:01 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 0x206a6c8>, <kernel.DependentProduct object at 0x24a4830>) of role type named in_type
% Using role type
% Declaring in:(fofType->(fofType->Prop))
% FOF formula (<kernel.Constant object at 0x206a488>, <kernel.DependentProduct object at 0x24a4830>) of role type named subset_type
% Using role type
% Declaring subset:(fofType->(fofType->Prop))
% FOF formula (((eq (fofType->(fofType->Prop))) subset) (fun (A:fofType) (B:fofType)=> (forall (Xx:fofType), (((in Xx) A)->((in Xx) B))))) of role definition named subset
% A new definition: (((eq (fofType->(fofType->Prop))) subset) (fun (A:fofType) (B:fofType)=> (forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))))
% Defined: subset:=(fun (A:fofType) (B:fofType)=> (forall (Xx:fofType), (((in Xx) A)->((in Xx) B))))
% FOF formula (<kernel.Constant object at 0x206a6c8>, <kernel.Sort object at 0x1f2f3f8>) of role type named subsetI1_type
% Using role type
% Declaring subsetI1:Prop
% FOF formula (((eq Prop) subsetI1) (forall (A:fofType) (B:fofType), ((forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))->((subset A) B)))) of role definition named subsetI1
% A new definition: (((eq Prop) subsetI1) (forall (A:fofType) (B:fofType), ((forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))->((subset A) B))))
% Defined: subsetI1:=(forall (A:fofType) (B:fofType), ((forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))->((subset A) B)))
% FOF formula (subsetI1->(forall (A:fofType) (B:fofType), ((forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))->((subset A) B)))) of role conjecture named subsetI2
% Conjecture to prove = (subsetI1->(forall (A:fofType) (B:fofType), ((forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))->((subset A) B)))):Prop
% Parameter fofType_DUMMY:fofType.
% We need to prove ['(subsetI1->(forall (A:fofType) (B:fofType), ((forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))->((subset A) B))))']
% Parameter fofType:Type.
% Parameter in:(fofType->(fofType->Prop)).
% Definition subset:=(fun (A:fofType) (B:fofType)=> (forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))):(fofType->(fofType->Prop)).
% Definition subsetI1:=(forall (A:fofType) (B:fofType), ((forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))->((subset A) B))):Prop.
% Trying to prove (subsetI1->(forall (A:fofType) (B:fofType), ((forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))->((subset A) B))))
% Found x:subsetI1
% Found (fun (x:subsetI1)=> x) as proof of (forall (A:fofType) (B:fofType), ((forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))->((subset A) B)))
% Found (fun (x:subsetI1)=> x) as proof of (subsetI1->(forall (A:fofType) (B:fofType), ((forall (Xx:fofType), (((in Xx) A)->((in Xx) B)))->((subset A) B))))
% Got proof (fun (x:subsetI1)=> x)
% Time elapsed = 0.032016s
% node=1 cost=3.000000 depth=1
% ::::::::::::::::::::::
% % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p
% % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% (fun (x:subsetI1)=> x)
% % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% EOF
%------------------------------------------------------------------------------