TSTP Solution File: NUM709^1 by cocATP---0.2.0
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% File : cocATP---0.2.0
% Problem : NUM709^1 : TPTP v7.0.0. Released v3.7.0.
% Transfm : none
% Format : tptp:raw
% Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p
% Computer : n154.star.cs.uiowa.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2609 0 2.40GHz
% Memory : 32218.625MB
% OS : Linux 3.10.0-693.2.2.el7.x86_64
% CPULimit : 300s
% DateTime : Mon Jan 8 13:11:32 EST 2018
% Result : Theorem 0.50s
% Output : Proof 0.50s
% Verified :
% SZS Type : None (Parsing solution fails)
% Syntax : Number of formulae : 0
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.03 % Problem : NUM709^1 : TPTP v7.0.0. Released v3.7.0.
% 0.00/0.03 % Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.02/0.23 % Computer : n154.star.cs.uiowa.edu
% 0.02/0.23 % Model : x86_64 x86_64
% 0.02/0.23 % CPU : Intel(R) Xeon(R) CPU E5-2609 0 @ 2.40GHz
% 0.02/0.23 % Memory : 32218.625MB
% 0.02/0.23 % OS : Linux 3.10.0-693.2.2.el7.x86_64
% 0.02/0.23 % CPULimit : 300
% 0.02/0.23 % DateTime : Fri Jan 5 13:10:34 CST 2018
% 0.02/0.23 % CPUTime :
% 0.07/0.27 Python 2.7.13
% 0.50/0.90 Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox2/benchmark/', '/export/starexec/sandbox2/benchmark/']
% 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59098>, <kernel.Type object at 0x2ae25ae59b48>) of role type named nat_type
% 0.50/0.90 Using role type
% 0.50/0.90 Declaring nat:Type
% 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59710>, <kernel.Constant object at 0x2ae25ae59ea8>) of role type named x
% 0.50/0.90 Using role type
% 0.50/0.90 Declaring x:nat
% 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59bd8>, <kernel.Constant object at 0x2ae25ae59098>) of role type named y
% 0.50/0.90 Using role type
% 0.50/0.90 Declaring y:nat
% 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59710>, <kernel.DependentProduct object at 0x2ae25aa8e0e0>) of role type named pl
% 0.50/0.90 Using role type
% 0.50/0.90 Declaring pl:(nat->(nat->nat))
% 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59098>, <kernel.DependentProduct object at 0x2ae25aa8ee60>) of role type named ts
% 0.50/0.90 Using role type
% 0.50/0.90 Declaring ts:(nat->(nat->nat))
% 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59710>, <kernel.DependentProduct object at 0x2ae25aa8eb00>) of role type named suc
% 0.50/0.90 Using role type
% 0.50/0.90 Declaring suc:(nat->nat)
% 0.50/0.90 FOF formula (forall (Xx:nat) (Xy:nat), (((eq nat) ((ts (suc Xx)) Xy)) ((pl ((ts Xx) Xy)) Xy))) of role axiom named satz28d
% 0.50/0.90 A new axiom: (forall (Xx:nat) (Xy:nat), (((eq nat) ((ts (suc Xx)) Xy)) ((pl ((ts Xx) Xy)) Xy)))
% 0.50/0.90 FOF formula (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y)) of role conjecture named satz28h
% 0.50/0.90 Conjecture to prove = (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y)):Prop
% 0.50/0.90 We need to prove ['(((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y))']
% 0.50/0.90 Parameter nat:Type.
% 0.50/0.90 Parameter x:nat.
% 0.50/0.90 Parameter y:nat.
% 0.50/0.90 Parameter pl:(nat->(nat->nat)).
% 0.50/0.90 Parameter ts:(nat->(nat->nat)).
% 0.50/0.90 Parameter suc:(nat->nat).
% 0.50/0.90 Axiom satz28d:(forall (Xx:nat) (Xy:nat), (((eq nat) ((ts (suc Xx)) Xy)) ((pl ((ts Xx) Xy)) Xy))).
% 0.50/0.90 Trying to prove (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y))
% 0.50/0.90 Found eq_ref00:=(eq_ref0 ((ts (suc x)) y)):(((eq nat) ((ts (suc x)) y)) ((ts (suc x)) y))
% 0.50/0.90 Found (eq_ref0 ((ts (suc x)) y)) as proof of (((eq nat) ((ts (suc x)) y)) ((ts (suc x)) y))
% 0.50/0.90 Found ((eq_ref nat) ((ts (suc x)) y)) as proof of (((eq nat) ((ts (suc x)) y)) ((ts (suc x)) y))
% 0.50/0.90 Found ((eq_ref nat) ((ts (suc x)) y)) as proof of (((eq nat) ((ts (suc x)) y)) ((ts (suc x)) y))
% 0.50/0.90 Found (satz28d000 ((eq_ref nat) ((ts (suc x)) y))) as proof of (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y))
% 0.50/0.90 Found ((satz28d00 (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y))) as proof of (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y))
% 0.50/0.90 Found (((satz28d0 y) (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y))) as proof of (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y))
% 0.50/0.90 Found ((((satz28d x) y) (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y))) as proof of (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y))
% 0.50/0.90 Found ((((satz28d x) y) (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y))) as proof of (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y))
% 0.50/0.90 Got proof ((((satz28d x) y) (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y)))
% 0.50/0.90 Time elapsed = 0.176284s
% 0.50/0.90 node=27 cost=-103.000000 depth=7
% 0.50/0.90::::::::::::::::::::::
% 0.50/0.90 % SZS status Theorem for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.50/0.90 % SZS output start Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.50/0.90 ((((satz28d x) y) (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y)))
% 0.50/0.90 % SZS output end Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p
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