TSTP Solution File: NUM639^1 by cocATP---0.2.0
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%------------------------------------------------------------------------------
% File : cocATP---0.2.0
% Problem : NUM639^1 : TPTP v7.0.0. Released v3.7.0.
% Transfm : none
% Format : tptp:raw
% Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p
% Computer : n185.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 0.47s
% Output : Proof 0.47s
% 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 : NUM639^1 : 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 : n185.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 11:19:00 CST 2018
% 0.02/0.23 % CPUTime :
% 0.06/0.26 Python 2.7.13
% 0.47/0.85 Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox2/benchmark/', '/export/starexec/sandbox2/benchmark/']
% 0.47/0.85 FOF formula (<kernel.Constant object at 0x2ba21e7f88c0>, <kernel.Type object at 0x2ba21e8d9758>) of role type named nat_type
% 0.47/0.85 Using role type
% 0.47/0.85 Declaring nat:Type
% 0.47/0.85 FOF formula (<kernel.Constant object at 0x2ba21e8d56c8>, <kernel.Constant object at 0x2ba21e8d96c8>) of role type named x
% 0.47/0.85 Using role type
% 0.47/0.85 Declaring x:nat
% 0.47/0.85 FOF formula (<kernel.Constant object at 0x2ba21e7f88c0>, <kernel.DependentProduct object at 0x2ba21e8d97a0>) of role type named suc
% 0.47/0.85 Using role type
% 0.47/0.85 Declaring suc:(nat->nat)
% 0.47/0.85 FOF formula (<kernel.Constant object at 0x2ba21e7f88c0>, <kernel.DependentProduct object at 0x2ba21e8d9488>) of role type named pl
% 0.47/0.85 Using role type
% 0.47/0.85 Declaring pl:(nat->(nat->nat))
% 0.47/0.85 FOF formula (<kernel.Constant object at 0x2ba216f5c2d8>, <kernel.Constant object at 0x2ba21e8d9a28>) of role type named n_1
% 0.47/0.85 Using role type
% 0.47/0.85 Declaring n_1:nat
% 0.47/0.85 FOF formula (forall (Xx:nat), (((eq nat) ((pl Xx) n_1)) (suc Xx))) of role axiom named satz4a
% 0.47/0.85 A new axiom: (forall (Xx:nat), (((eq nat) ((pl Xx) n_1)) (suc Xx)))
% 0.47/0.85 FOF formula (((eq nat) (suc x)) ((pl x) n_1)) of role conjecture named satz4e
% 0.47/0.85 Conjecture to prove = (((eq nat) (suc x)) ((pl x) n_1)):Prop
% 0.47/0.85 We need to prove ['(((eq nat) (suc x)) ((pl x) n_1))']
% 0.47/0.85 Parameter nat:Type.
% 0.47/0.85 Parameter x:nat.
% 0.47/0.85 Parameter suc:(nat->nat).
% 0.47/0.85 Parameter pl:(nat->(nat->nat)).
% 0.47/0.85 Parameter n_1:nat.
% 0.47/0.85 Axiom satz4a:(forall (Xx:nat), (((eq nat) ((pl Xx) n_1)) (suc Xx))).
% 0.47/0.85 Trying to prove (((eq nat) (suc x)) ((pl x) n_1))
% 0.47/0.85 Found eq_ref00:=(eq_ref0 ((pl x) n_1)):(((eq nat) ((pl x) n_1)) ((pl x) n_1))
% 0.47/0.85 Found (eq_ref0 ((pl x) n_1)) as proof of (((eq nat) ((pl x) n_1)) ((pl x) n_1))
% 0.47/0.85 Found ((eq_ref nat) ((pl x) n_1)) as proof of (((eq nat) ((pl x) n_1)) ((pl x) n_1))
% 0.47/0.85 Found ((eq_ref nat) ((pl x) n_1)) as proof of (((eq nat) ((pl x) n_1)) ((pl x) n_1))
% 0.47/0.85 Found (satz4a00 ((eq_ref nat) ((pl x) n_1))) as proof of (((eq nat) (suc x)) ((pl x) n_1))
% 0.47/0.85 Found ((satz4a0 (fun (x1:nat)=> (((eq nat) x1) ((pl x) n_1)))) ((eq_ref nat) ((pl x) n_1))) as proof of (((eq nat) (suc x)) ((pl x) n_1))
% 0.47/0.85 Found (((satz4a x) (fun (x1:nat)=> (((eq nat) x1) ((pl x) n_1)))) ((eq_ref nat) ((pl x) n_1))) as proof of (((eq nat) (suc x)) ((pl x) n_1))
% 0.47/0.85 Found (((satz4a x) (fun (x1:nat)=> (((eq nat) x1) ((pl x) n_1)))) ((eq_ref nat) ((pl x) n_1))) as proof of (((eq nat) (suc x)) ((pl x) n_1))
% 0.47/0.85 Got proof (((satz4a x) (fun (x1:nat)=> (((eq nat) x1) ((pl x) n_1)))) ((eq_ref nat) ((pl x) n_1)))
% 0.47/0.85 Time elapsed = 0.147780s
% 0.47/0.85 node=25 cost=-130.000000 depth=6
% 0.47/0.85::::::::::::::::::::::
% 0.47/0.85 % SZS status Theorem for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.47/0.85 % SZS output start Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.47/0.85 (((satz4a x) (fun (x1:nat)=> (((eq nat) x1) ((pl x) n_1)))) ((eq_ref nat) ((pl x) n_1)))
% 0.47/0.85 % SZS output end Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p
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