TSTP Solution File: LDA001-1 by Matita---1.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Matita---1.0
% Problem : LDA001-1 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : matitaprover --timeout %d --tptppath /export/starexec/sandbox2/benchmark %s
% Computer : n019.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 : 600s
% DateTime : Sun Jul 17 16:44:55 EDT 2022
% Result : Unsatisfiable 0.19s 0.37s
% Output : CNFRefutation 0.19s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.12 % Problem : LDA001-1 : TPTP v8.1.0. Released v1.0.0.
% 0.04/0.13 % Command : matitaprover --timeout %d --tptppath /export/starexec/sandbox2/benchmark %s
% 0.12/0.34 % Computer : n019.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 600
% 0.12/0.34 % DateTime : Mon May 30 02:11:10 EDT 2022
% 0.12/0.34 % CPUTime :
% 0.12/0.34 15295: Facts:
% 0.12/0.34 15295: Id : 2, {_}:
% 0.12/0.34 f ?2 (f ?3 ?4) =<= f (f ?2 ?3) (f ?2 ?4)
% 0.12/0.34 [4, 3, 2] by a1 ?2 ?3 ?4
% 0.12/0.34 15295: Id : 3, {_}: n2 =<= f n1 n1 [] by clause_2
% 0.12/0.34 15295: Id : 4, {_}: n3 =<= f n2 n1 [] by clause_3
% 0.12/0.34 15295: Id : 5, {_}: u =<= f n2 n2 [] by clause_4
% 0.12/0.34 15295: Goal:
% 0.12/0.34 15295: Id : 1, {_}: f (f n3 n2) u =<= f (f u u) u [] by prove_equation
% 0.19/0.37 Statistics :
% 0.19/0.37 Max weight : 11
% 0.19/0.37 Found proof, 0.026945s
% 0.19/0.37 % SZS status Unsatisfiable for theBenchmark.p
% 0.19/0.37 % SZS output start CNFRefutation for theBenchmark.p
% 0.19/0.37 Id : 5, {_}: u =<= f n2 n2 [] by clause_4
% 0.19/0.37 Id : 3, {_}: n2 =<= f n1 n1 [] by clause_2
% 0.19/0.37 Id : 4, {_}: n3 =<= f n2 n1 [] by clause_3
% 0.19/0.37 Id : 2, {_}: f ?2 (f ?3 ?4) =<= f (f ?2 ?3) (f ?2 ?4) [4, 3, 2] by a1 ?2 ?3 ?4
% 0.19/0.37 Id : 87, {_}: f n2 (f ?129 n1) =<= f (f n2 ?129) n3 [129] by Super 2 with 4 at 2,3
% 0.19/0.37 Id : 88, {_}: f n2 (f n1 n1) =>= f n3 n3 [] by Super 87 with 4 at 1,3
% 0.19/0.37 Id : 91, {_}: f n2 n2 =>= f n3 n3 [] by Demod 88 with 3 at 2,2
% 0.19/0.37 Id : 92, {_}: u =<= f n3 n3 [] by Demod 91 with 5 at 2
% 0.19/0.37 Id : 94, {_}: f n3 (f ?135 n3) =<= f (f n3 ?135) u [135] by Super 2 with 92 at 2,3
% 0.19/0.37 Id : 150, {_}: f n2 (f ?200 n2) =<= f (f n2 ?200) u [200] by Super 2 with 5 at 2,3
% 0.19/0.37 Id : 156, {_}: f n2 (f n2 n2) =>= f u u [] by Super 150 with 5 at 1,3
% 0.19/0.37 Id : 167, {_}: f n2 u =<= f u u [] by Demod 156 with 5 at 2,2
% 0.19/0.37 Id : 106, {_}: f n2 (f n1 ?155) =>= f n3 (f n2 ?155) [155] by Super 2 with 4 at 1,3
% 0.19/0.37 Id : 24, {_}: f n1 (f ?51 n1) =<= f (f n1 ?51) n2 [51] by Super 2 with 3 at 2,3
% 0.19/0.37 Id : 25, {_}: f n1 (f n1 n1) =>= f n2 n2 [] by Super 24 with 3 at 1,3
% 0.19/0.37 Id : 27, {_}: f n1 n2 =>= f n2 n2 [] by Demod 25 with 3 at 2,2
% 0.19/0.37 Id : 28, {_}: f n1 n2 =>= u [] by Demod 27 with 5 at 3
% 0.19/0.37 Id : 107, {_}: f n2 u =<= f n3 (f n2 n2) [] by Super 106 with 28 at 2,2
% 0.19/0.37 Id : 115, {_}: f n2 u =>= f n3 u [] by Demod 107 with 5 at 2,3
% 0.19/0.37 Id : 168, {_}: f n3 u =<= f u u [] by Demod 167 with 115 at 2
% 0.19/0.37 Id : 89, {_}: f n2 (f n2 n1) =>= f u n3 [] by Super 87 with 5 at 1,3
% 0.19/0.37 Id : 93, {_}: f n2 n3 =<= f u n3 [] by Demod 89 with 4 at 2,2
% 0.19/0.37 Id : 282, {_}: f n3 (f n2 n3) === f n3 (f n2 n3) [] by Demod 281 with 93 at 2,3
% 0.19/0.37 Id : 281, {_}: f n3 (f n2 n3) =<= f n3 (f u n3) [] by Demod 280 with 94 at 3
% 0.19/0.37 Id : 280, {_}: f n3 (f n2 n3) =<= f (f n3 u) u [] by Demod 279 with 168 at 1,3
% 0.19/0.37 Id : 279, {_}: f n3 (f n2 n3) =<= f (f u u) u [] by Demod 1 with 94 at 2
% 0.19/0.37 Id : 1, {_}: f (f n3 n2) u =<= f (f u u) u [] by prove_equation
% 0.19/0.37 % SZS output end CNFRefutation for theBenchmark.p
% 0.19/0.37 15298: solved /export/starexec/sandbox2/benchmark/theBenchmark.p in 0.028341 using nrkbo
%------------------------------------------------------------------------------