TSTP Solution File: SYO226^5 by cocATP---0.2.0
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- Process Solution
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% File : cocATP---0.2.0
% Problem : SYO226^5 : TPTP v7.5.0. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% Computer : n020.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 00:50:57 EDT 2022
% Result : Theorem 0.40s 0.61s
% Output : Proof 0.40s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.11 % Problem : SYO226^5 : TPTP v7.5.0. Released v4.0.0.
% 0.07/0.12 % Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.12/0.33 % Computer : n020.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 20:02:41 EST 2022
% 0.12/0.33 % CPUTime :
% 0.12/0.33 ModuleCmd_Load.c(213):ERROR:105: Unable to locate a modulefile for 'python/python27'
% 0.12/0.34 Python 2.7.5
% 0.40/0.61 Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox/benchmark/', '/export/starexec/sandbox/benchmark/']
% 0.40/0.61 FOF formula (forall (X:fofType) (Y:fofType), ((forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y)))->(((eq fofType) X) Y))) of role conjecture named cTHM47B
% 0.40/0.61 Conjecture to prove = (forall (X:fofType) (Y:fofType), ((forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y)))->(((eq fofType) X) Y))):Prop
% 0.40/0.61 Parameter fofType_DUMMY:fofType.
% 0.40/0.61 We need to prove ['(forall (X:fofType) (Y:fofType), ((forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y)))->(((eq fofType) X) Y)))']
% 0.40/0.61 Parameter fofType:Type.
% 0.40/0.61 Trying to prove (forall (X:fofType) (Y:fofType), ((forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y)))->(((eq fofType) X) Y)))
% 0.40/0.61 Found eq_ref0:=(eq_ref fofType):(forall (a:fofType), (((eq fofType) a) a))
% 0.40/0.61 Found (eq_ref fofType) as proof of (forall (Z:fofType), (((eq fofType) Z) Z))
% 0.40/0.61 Found (eq_ref fofType) as proof of (forall (Z:fofType), (((eq fofType) Z) Z))
% 0.40/0.61 Found (x0 (eq_ref fofType)) as proof of (((eq fofType) X) Y)
% 0.40/0.61 Found ((x (eq fofType)) (eq_ref fofType)) as proof of (((eq fofType) X) Y)
% 0.40/0.61 Found (fun (x:(forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y))))=> ((x (eq fofType)) (eq_ref fofType))) as proof of (((eq fofType) X) Y)
% 0.40/0.61 Found (fun (Y:fofType) (x:(forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y))))=> ((x (eq fofType)) (eq_ref fofType))) as proof of ((forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y)))->(((eq fofType) X) Y))
% 0.40/0.61 Found (fun (X:fofType) (Y:fofType) (x:(forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y))))=> ((x (eq fofType)) (eq_ref fofType))) as proof of (forall (Y:fofType), ((forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y)))->(((eq fofType) X) Y)))
% 0.40/0.61 Found (fun (X:fofType) (Y:fofType) (x:(forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y))))=> ((x (eq fofType)) (eq_ref fofType))) as proof of (forall (X:fofType) (Y:fofType), ((forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y)))->(((eq fofType) X) Y)))
% 0.40/0.61 Got proof (fun (X:fofType) (Y:fofType) (x:(forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y))))=> ((x (eq fofType)) (eq_ref fofType)))
% 0.40/0.61 Time elapsed = 0.107241s
% 0.40/0.61 node=19 cost=-46.000000 depth=7
% 0.40/0.61 ::::::::::::::::::::::
% 0.40/0.61 % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.40/0.61 % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.40/0.61 (fun (X:fofType) (Y:fofType) (x:(forall (R:(fofType->(fofType->Prop))), ((forall (Z:fofType), ((R Z) Z))->((R X) Y))))=> ((x (eq fofType)) (eq_ref fofType)))
% 0.40/0.61 % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
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