TSTP Solution File: NUM669^1 by cocATP---0.2.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : cocATP---0.2.0
% Problem : NUM669^1 : TPTP v7.0.0. Released v3.7.0.
% Transfm : none
% Format : tptp:raw
% Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% Computer : n060.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 13:11:22 EST 2018
% Result : Theorem 0.09s
% Output : Proof 0.09s
% Verified :
% SZS Type : None (Parsing solution fails)
% Syntax : Number of formulae : 0
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.03 % Problem : NUM669^1 : TPTP v7.0.0. Released v3.7.0.
% 0.00/0.03 % Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.03/0.23 % Computer : n060.star.cs.uiowa.edu
% 0.03/0.23 % Model : x86_64 x86_64
% 0.03/0.23 % CPU : Intel(R) Xeon(R) CPU E5-2609 0 @ 2.40GHz
% 0.03/0.23 % Memory : 32218.625MB
% 0.03/0.23 % OS : Linux 3.10.0-693.2.2.el7.x86_64
% 0.03/0.23 % CPULimit : 300
% 0.03/0.23 % DateTime : Fri Jan 5 12:13:49 CST 2018
% 0.03/0.23 % CPUTime :
% 0.03/0.25 Python 2.7.13
% 0.09/0.56 Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox/benchmark/', '/export/starexec/sandbox/benchmark/']
% 0.09/0.56 FOF formula (<kernel.Constant object at 0x2ab07800d3f8>, <kernel.Type object at 0x2ab07800de60>) of role type named nat_type
% 0.09/0.56 Using role type
% 0.09/0.56 Declaring nat:Type
% 0.09/0.56 FOF formula (<kernel.Constant object at 0x2ab078704ea8>, <kernel.Constant object at 0x2ab07800df80>) of role type named x
% 0.09/0.56 Using role type
% 0.09/0.56 Declaring x:nat
% 0.09/0.56 FOF formula (<kernel.Constant object at 0x2ab07800d3b0>, <kernel.DependentProduct object at 0x2ab07800dcb0>) of role type named more
% 0.09/0.56 Using role type
% 0.09/0.56 Declaring more:(nat->(nat->Prop))
% 0.09/0.56 FOF formula (<kernel.Constant object at 0x2ab07800d5f0>, <kernel.DependentProduct object at 0x2ab07800da28>) of role type named suc
% 0.09/0.56 Using role type
% 0.09/0.56 Declaring suc:(nat->nat)
% 0.09/0.56 FOF formula (<kernel.Constant object at 0x2ab07800df80>, <kernel.DependentProduct object at 0x2ab07800dfc8>) of role type named pl
% 0.09/0.56 Using role type
% 0.09/0.56 Declaring pl:(nat->(nat->nat))
% 0.09/0.56 FOF formula (<kernel.Constant object at 0x2ab07800dcb0>, <kernel.Constant object at 0x2ab07800dfc8>) of role type named n_1
% 0.09/0.56 Using role type
% 0.09/0.56 Declaring n_1:nat
% 0.09/0.56 FOF formula (forall (Xx:nat) (Xy:nat), ((more ((pl Xx) Xy)) Xx)) of role axiom named satz18
% 0.09/0.56 A new axiom: (forall (Xx:nat) (Xy:nat), ((more ((pl Xx) Xy)) Xx))
% 0.09/0.56 FOF formula (forall (Xx:nat), (((eq nat) ((pl Xx) n_1)) (suc Xx))) of role axiom named satz4a
% 0.09/0.56 A new axiom: (forall (Xx:nat), (((eq nat) ((pl Xx) n_1)) (suc Xx)))
% 0.09/0.56 FOF formula ((more (suc x)) x) of role conjecture named satz18b
% 0.09/0.56 Conjecture to prove = ((more (suc x)) x):Prop
% 0.09/0.56 We need to prove ['((more (suc x)) x)']
% 0.09/0.56 Parameter nat:Type.
% 0.09/0.56 Parameter x:nat.
% 0.09/0.56 Parameter more:(nat->(nat->Prop)).
% 0.09/0.56 Parameter suc:(nat->nat).
% 0.09/0.56 Parameter pl:(nat->(nat->nat)).
% 0.09/0.56 Parameter n_1:nat.
% 0.09/0.56 Axiom satz18:(forall (Xx:nat) (Xy:nat), ((more ((pl Xx) Xy)) Xx)).
% 0.09/0.56 Axiom satz4a:(forall (Xx:nat), (((eq nat) ((pl Xx) n_1)) (suc Xx))).
% 0.09/0.56 Trying to prove ((more (suc x)) x)
% 0.09/0.56 Found satz1800:=(satz180 n_1):((more ((pl x) n_1)) x)
% 0.09/0.56 Found (satz180 n_1) as proof of ((more ((pl x) n_1)) x)
% 0.09/0.56 Found ((satz18 x) n_1) as proof of ((more ((pl x) n_1)) x)
% 0.09/0.56 Found ((satz18 x) n_1) as proof of ((more ((pl x) n_1)) x)
% 0.09/0.56 Found (satz4a00 ((satz18 x) n_1)) as proof of ((more (suc x)) x)
% 0.09/0.56 Found ((satz4a0 (fun (x1:nat)=> ((more x1) x))) ((satz18 x) n_1)) as proof of ((more (suc x)) x)
% 0.09/0.56 Found (((satz4a x) (fun (x1:nat)=> ((more x1) x))) ((satz18 x) n_1)) as proof of ((more (suc x)) x)
% 0.09/0.56 Found (((satz4a x) (fun (x1:nat)=> ((more x1) x))) ((satz18 x) n_1)) as proof of ((more (suc x)) x)
% 0.09/0.56 Got proof (((satz4a x) (fun (x1:nat)=> ((more x1) x))) ((satz18 x) n_1))
% 0.09/0.56 Time elapsed = 0.041541s
% 0.09/0.56 node=11 cost=-136.000000 depth=6
% 0.09/0.56::::::::::::::::::::::
% 0.09/0.56 % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.09/0.56 % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.09/0.56 (((satz4a x) (fun (x1:nat)=> ((more x1) x))) ((satz18 x) n_1))
% 0.09/0.56 % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
%------------------------------------------------------------------------------