TSTP Solution File: SYO356^5 by cocATP---0.2.0
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- Process Solution
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% File : cocATP---0.2.0
% Problem : SYO356^5 : TPTP v7.5.0. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p
% Computer : n029.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 00:51:14 EDT 2022
% Result : Theorem 0.54s 0.72s
% Output : Proof 0.54s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.10 % Problem : SYO356^5 : TPTP v7.5.0. Released v4.0.0.
% 0.06/0.11 % Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.11/0.32 % Computer : n029.cluster.edu
% 0.11/0.32 % Model : x86_64 x86_64
% 0.11/0.32 % CPUModel : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.32 % RAMPerCPU : 8042.1875MB
% 0.11/0.32 % OS : Linux 3.10.0-693.el7.x86_64
% 0.11/0.32 % CPULimit : 300
% 0.11/0.32 % DateTime : Sat Mar 12 06:48:45 EST 2022
% 0.11/0.32 % CPUTime :
% 0.11/0.33 ModuleCmd_Load.c(213):ERROR:105: Unable to locate a modulefile for 'python/python27'
% 0.11/0.33 Python 2.7.5
% 0.54/0.72 Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox2/benchmark/', '/export/starexec/sandbox2/benchmark/']
% 0.54/0.72 FOF formula (<kernel.Constant object at 0x1f124d0>, <kernel.Type object at 0x21a6dd0>) of role type named a_type
% 0.54/0.72 Using role type
% 0.54/0.72 Declaring a:Type
% 0.54/0.72 FOF formula (<kernel.Constant object at 0x1f08e60>, <kernel.Constant object at 0x21a6908>) of role type named v
% 0.54/0.72 Using role type
% 0.54/0.72 Declaring v:a
% 0.54/0.72 FOF formula (<kernel.Constant object at 0x2b9b116ba950>, <kernel.Constant object at 0x21a6908>) of role type named u
% 0.54/0.72 Using role type
% 0.54/0.72 Declaring u:a
% 0.54/0.72 FOF formula ((forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v)))->(forall (Xq:(a->Prop)), ((Xq u)->(Xq v)))) of role conjecture named cE1LEIBEQ2_pme
% 0.54/0.72 Conjecture to prove = ((forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v)))->(forall (Xq:(a->Prop)), ((Xq u)->(Xq v)))):Prop
% 0.54/0.72 We need to prove ['((forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v)))->(forall (Xq:(a->Prop)), ((Xq u)->(Xq v))))']
% 0.54/0.72 Parameter a:Type.
% 0.54/0.72 Parameter v:a.
% 0.54/0.72 Parameter u:a.
% 0.54/0.72 Trying to prove ((forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v)))->(forall (Xq:(a->Prop)), ((Xq u)->(Xq v))))
% 0.54/0.72 Found eq_ref0:=(eq_ref a):(forall (a0:a), (((eq a) a0) a0))
% 0.54/0.72 Found (eq_ref a) as proof of (forall (Z:a), (((eq a) Z) Z))
% 0.54/0.72 Found (eq_ref a) as proof of (forall (Z:a), (((eq a) Z) Z))
% 0.54/0.72 Found (x0 (eq_ref a)) as proof of (((eq a) u) v)
% 0.54/0.72 Found ((x (eq a)) (eq_ref a)) as proof of (((eq a) u) v)
% 0.54/0.72 Found (fun (x:(forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v))))=> ((x (eq a)) (eq_ref a))) as proof of (((eq a) u) v)
% 0.54/0.72 Found (fun (x:(forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v))))=> ((x (eq a)) (eq_ref a))) as proof of ((forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v)))->(((eq a) u) v))
% 0.54/0.72 Found (fun (x:(forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v))))=> ((x (eq a)) (eq_ref a))) as proof of ((forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v)))->(forall (Xq:(a->Prop)), ((Xq u)->(Xq v))))
% 0.54/0.72 Got proof (fun (x:(forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v))))=> ((x (eq a)) (eq_ref a)))
% 0.54/0.72 Time elapsed = 0.120294s
% 0.54/0.72 node=25 cost=-11.000000 depth=6
% 0.54/0.72 ::::::::::::::::::::::
% 0.54/0.72 % SZS status Theorem for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.54/0.72 % SZS output start Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.54/0.72 (fun (x:(forall (Q:(a->(a->Prop))), ((forall (Z:a), ((Q Z) Z))->((Q u) v))))=> ((x (eq a)) (eq_ref a)))
% 0.54/0.72 % SZS output end Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p
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