TSTP Solution File: NUM642^1 by cocATP---0.2.0
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% File : cocATP---0.2.0
% Problem : NUM642^1 : TPTP v7.0.0. Released v3.7.0.
% Transfm : none
% Format : tptp:raw
% Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p
% Computer : n119.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 13:11:15 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
% 0.00/0.04 % Problem : NUM642^1 : TPTP v7.0.0. Released v3.7.0.
% 0.00/0.04 % Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.02/0.23 % Computer : n119.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 11:24:00 CST 2018
% 0.02/0.23 % CPUTime :
% 0.02/0.26 Python 2.7.13
% 0.50/0.88 Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox2/benchmark/', '/export/starexec/sandbox2/benchmark/']
% 0.50/0.88 FOF formula (<kernel.Constant object at 0x2b8e1154a440>, <kernel.Type object at 0x2b8e1154a8c0>) of role type named nat_type
% 0.50/0.88 Using role type
% 0.50/0.88 Declaring nat:Type
% 0.50/0.88 FOF formula (<kernel.Constant object at 0x2b8e119a7ea8>, <kernel.Constant object at 0x2b8e1154a680>) of role type named x
% 0.50/0.88 Using role type
% 0.50/0.88 Declaring x:nat
% 0.50/0.88 FOF formula (<kernel.Constant object at 0x2b8e1154a7e8>, <kernel.Constant object at 0x2b8e1154a680>) of role type named y
% 0.50/0.88 Using role type
% 0.50/0.88 Declaring y:nat
% 0.50/0.88 FOF formula (<kernel.Constant object at 0x2b8e1154a440>, <kernel.DependentProduct object at 0x2b8e11919680>) of role type named suc
% 0.50/0.88 Using role type
% 0.50/0.88 Declaring suc:(nat->nat)
% 0.50/0.88 FOF formula (<kernel.Constant object at 0x2b8e1154a680>, <kernel.DependentProduct object at 0x2b8e119190e0>) of role type named pl
% 0.50/0.88 Using role type
% 0.50/0.88 Declaring pl:(nat->(nat->nat))
% 0.50/0.88 FOF formula (forall (Xx:nat) (Xy:nat), (((eq nat) ((pl (suc Xx)) Xy)) (suc ((pl Xx) Xy)))) of role axiom named satz4d
% 0.50/0.88 A new axiom: (forall (Xx:nat) (Xy:nat), (((eq nat) ((pl (suc Xx)) Xy)) (suc ((pl Xx) Xy))))
% 0.50/0.88 FOF formula (((eq nat) (suc ((pl x) y))) ((pl (suc x)) y)) of role conjecture named satz4h
% 0.50/0.88 Conjecture to prove = (((eq nat) (suc ((pl x) y))) ((pl (suc x)) y)):Prop
% 0.50/0.88 We need to prove ['(((eq nat) (suc ((pl x) y))) ((pl (suc x)) y))']
% 0.50/0.88 Parameter nat:Type.
% 0.50/0.88 Parameter x:nat.
% 0.50/0.88 Parameter y:nat.
% 0.50/0.88 Parameter suc:(nat->nat).
% 0.50/0.88 Parameter pl:(nat->(nat->nat)).
% 0.50/0.88 Axiom satz4d:(forall (Xx:nat) (Xy:nat), (((eq nat) ((pl (suc Xx)) Xy)) (suc ((pl Xx) Xy)))).
% 0.50/0.88 Trying to prove (((eq nat) (suc ((pl x) y))) ((pl (suc x)) y))
% 0.50/0.88 Found eq_ref00:=(eq_ref0 ((pl (suc x)) y)):(((eq nat) ((pl (suc x)) y)) ((pl (suc x)) y))
% 0.50/0.88 Found (eq_ref0 ((pl (suc x)) y)) as proof of (((eq nat) ((pl (suc x)) y)) ((pl (suc x)) y))
% 0.50/0.88 Found ((eq_ref nat) ((pl (suc x)) y)) as proof of (((eq nat) ((pl (suc x)) y)) ((pl (suc x)) y))
% 0.50/0.88 Found ((eq_ref nat) ((pl (suc x)) y)) as proof of (((eq nat) ((pl (suc x)) y)) ((pl (suc x)) y))
% 0.50/0.88 Found (satz4d000 ((eq_ref nat) ((pl (suc x)) y))) as proof of (((eq nat) (suc ((pl x) y))) ((pl (suc x)) y))
% 0.50/0.88 Found ((satz4d00 (fun (x1:nat)=> (((eq nat) x1) ((pl (suc x)) y)))) ((eq_ref nat) ((pl (suc x)) y))) as proof of (((eq nat) (suc ((pl x) y))) ((pl (suc x)) y))
% 0.50/0.88 Found (((satz4d0 y) (fun (x1:nat)=> (((eq nat) x1) ((pl (suc x)) y)))) ((eq_ref nat) ((pl (suc x)) y))) as proof of (((eq nat) (suc ((pl x) y))) ((pl (suc x)) y))
% 0.50/0.88 Found ((((satz4d x) y) (fun (x1:nat)=> (((eq nat) x1) ((pl (suc x)) y)))) ((eq_ref nat) ((pl (suc x)) y))) as proof of (((eq nat) (suc ((pl x) y))) ((pl (suc x)) y))
% 0.50/0.88 Found ((((satz4d x) y) (fun (x1:nat)=> (((eq nat) x1) ((pl (suc x)) y)))) ((eq_ref nat) ((pl (suc x)) y))) as proof of (((eq nat) (suc ((pl x) y))) ((pl (suc x)) y))
% 0.50/0.88 Got proof ((((satz4d x) y) (fun (x1:nat)=> (((eq nat) x1) ((pl (suc x)) y)))) ((eq_ref nat) ((pl (suc x)) y)))
% 0.50/0.88 Time elapsed = 0.170933s
% 0.50/0.88 node=27 cost=-103.000000 depth=7
% 0.50/0.88::::::::::::::::::::::
% 0.50/0.88 % SZS status Theorem for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.50/0.88 % SZS output start Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.50/0.88 ((((satz4d x) y) (fun (x1:nat)=> (((eq nat) x1) ((pl (suc x)) y)))) ((eq_ref nat) ((pl (suc x)) y)))
% 0.50/0.88 % SZS output end Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p
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