TSTP Solution File: SYO141^5 by cocATP---0.2.0
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%------------------------------------------------------------------------------
% File : cocATP---0.2.0
% Problem : SYO141^5 : TPTP v7.5.0. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p
% Computer : n004.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit : 0s
% DateTime : Tue Mar 29 00:50:44 EDT 2022
% Result : Theorem 0.46s 0.65s
% Output : Proof 0.46s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.11 % Problem : SYO141^5 : TPTP v7.5.0. Released v4.0.0.
% 0.11/0.12 % Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.12/0.33 % Computer : n004.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPUModel : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % RAMPerCPU : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % DateTime : Fri Mar 11 16:21:25 EST 2022
% 0.12/0.33 % CPUTime :
% 0.12/0.34 ModuleCmd_Load.c(213):ERROR:105: Unable to locate a modulefile for 'python/python27'
% 0.12/0.34 Python 2.7.5
% 0.46/0.65 Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox2/benchmark/', '/export/starexec/sandbox2/benchmark/']
% 0.46/0.65 FOF formula (<kernel.Constant object at 0xe7c3f8>, <kernel.Constant object at 0xe7c7e8>) of role type named a
% 0.46/0.65 Using role type
% 0.46/0.65 Declaring a:fofType
% 0.46/0.65 FOF formula (<kernel.Constant object at 0xe76e18>, <kernel.DependentProduct object at 0xe7cf80>) of role type named cQ
% 0.46/0.65 Using role type
% 0.46/0.65 Declaring cQ:(fofType->Prop)
% 0.46/0.65 FOF formula (<kernel.Constant object at 0xe7c3f8>, <kernel.Single object at 0xe7c128>) of role type named x
% 0.46/0.65 Using role type
% 0.46/0.65 Declaring x:fofType
% 0.46/0.65 FOF formula (<kernel.Constant object at 0xe7c368>, <kernel.DependentProduct object at 0xe7c6c8>) of role type named cR
% 0.46/0.65 Using role type
% 0.46/0.65 Declaring cR:(fofType->Prop)
% 0.46/0.65 FOF formula (<kernel.Constant object at 0xe7c128>, <kernel.DependentProduct object at 0x2b7fa14e1fc8>) of role type named cP
% 0.46/0.65 Using role type
% 0.46/0.65 Declaring cP:(fofType->Prop)
% 0.46/0.65 FOF formula (((cP a)->(forall (Xx0:fofType), (cQ Xx0)))->((forall (Xx0:fofType), (cP Xx0))->((cR x)->(cQ a)))) of role conjecture named cADDHYP2
% 0.46/0.65 Conjecture to prove = (((cP a)->(forall (Xx0:fofType), (cQ Xx0)))->((forall (Xx0:fofType), (cP Xx0))->((cR x)->(cQ a)))):Prop
% 0.46/0.65 We need to prove ['(((cP a)->(forall (Xx0:fofType), (cQ Xx0)))->((forall (Xx0:fofType), (cP Xx0))->((cR x)->(cQ a))))']
% 0.46/0.65 Parameter fofType:Type.
% 0.46/0.65 Parameter a:fofType.
% 0.46/0.65 Parameter cQ:(fofType->Prop).
% 0.46/0.65 Parameter x:fofType.
% 0.46/0.65 Parameter cR:(fofType->Prop).
% 0.46/0.65 Parameter cP:(fofType->Prop).
% 0.46/0.65 Trying to prove (((cP a)->(forall (Xx0:fofType), (cQ Xx0)))->((forall (Xx0:fofType), (cP Xx0))->((cR x)->(cQ a))))
% 0.46/0.65 Found x000:=(x00 a):(cP a)
% 0.46/0.65 Found (x00 a) as proof of (cP a)
% 0.46/0.65 Found (x00 a) as proof of (cP a)
% 0.46/0.65 Found (x01 (x00 a)) as proof of (cQ a)
% 0.46/0.65 Found ((fun (x2:(cP a))=> ((x0 x2) a)) (x00 a)) as proof of (cQ a)
% 0.46/0.65 Found (fun (x1:(cR x))=> ((fun (x2:(cP a))=> ((x0 x2) a)) (x00 a))) as proof of (cQ a)
% 0.46/0.65 Found (fun (x00:(forall (Xx0:fofType), (cP Xx0))) (x1:(cR x))=> ((fun (x2:(cP a))=> ((x0 x2) a)) (x00 a))) as proof of ((cR x)->(cQ a))
% 0.46/0.65 Found (fun (x0:((cP a)->(forall (Xx0:fofType), (cQ Xx0)))) (x00:(forall (Xx0:fofType), (cP Xx0))) (x1:(cR x))=> ((fun (x2:(cP a))=> ((x0 x2) a)) (x00 a))) as proof of ((forall (Xx0:fofType), (cP Xx0))->((cR x)->(cQ a)))
% 0.46/0.65 Found (fun (x0:((cP a)->(forall (Xx0:fofType), (cQ Xx0)))) (x00:(forall (Xx0:fofType), (cP Xx0))) (x1:(cR x))=> ((fun (x2:(cP a))=> ((x0 x2) a)) (x00 a))) as proof of (((cP a)->(forall (Xx0:fofType), (cQ Xx0)))->((forall (Xx0:fofType), (cP Xx0))->((cR x)->(cQ a))))
% 0.46/0.65 Got proof (fun (x0:((cP a)->(forall (Xx0:fofType), (cQ Xx0)))) (x00:(forall (Xx0:fofType), (cP Xx0))) (x1:(cR x))=> ((fun (x2:(cP a))=> ((x0 x2) a)) (x00 a)))
% 0.46/0.65 Time elapsed = 0.035547s
% 0.46/0.65 node=10 cost=30.000000 depth=7
% 0.46/0.65 ::::::::::::::::::::::
% 0.46/0.65 % SZS status Theorem for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.46/0.65 % SZS output start Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.46/0.65 (fun (x0:((cP a)->(forall (Xx0:fofType), (cQ Xx0)))) (x00:(forall (Xx0:fofType), (cP Xx0))) (x1:(cR x))=> ((fun (x2:(cP a))=> ((x0 x2) a)) (x00 a)))
% 0.46/0.65 % SZS output end Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p
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