TSTP Solution File: NUM670^1 by Duper---1.0
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% File : Duper---1.0
% Problem : NUM670^1 : TPTP v8.1.2. Released v3.7.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n022.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 : 300s
% DateTime : Thu Aug 31 10:56:56 EDT 2023
% Result : Theorem 3.42s 3.79s
% Output : Proof 3.42s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : NUM670^1 : TPTP v8.1.2. Released v3.7.0.
% 0.00/0.14 % Command : duper %s
% 0.13/0.35 % Computer : n022.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Fri Aug 25 13:41:25 EDT 2023
% 0.13/0.35 % CPUTime :
% 3.42/3.79 SZS status Theorem for theBenchmark.p
% 3.42/3.79 SZS output start Proof for theBenchmark.p
% 3.42/3.79 Clause #0 (by assumption #[]): Eq (Not (∀ (Xx_0 : nat), Ne x (pl y Xx_0))) True
% 3.42/3.79 Clause #2 (by assumption #[]): Eq (∀ (Xx Xy : nat), Eq (pl Xx Xy) (pl Xy Xx)) True
% 3.42/3.79 Clause #3 (by assumption #[]): Eq (∀ (Xx Xy Xz : nat), Eq (pl (pl Xx Xy) Xz) (pl Xx (pl Xy Xz))) True
% 3.42/3.79 Clause #4 (by assumption #[]): Eq (Not (Not (∀ (Xx_0 : nat), Ne (pl x z) (pl (pl y z) Xx_0)))) True
% 3.42/3.79 Clause #9 (by clausification #[0]): Eq (∀ (Xx_0 : nat), Ne x (pl y Xx_0)) False
% 3.42/3.79 Clause #10 (by clausification #[9]): ∀ (a : nat), Eq (Not (Ne x (pl y (skS.0 0 a)))) True
% 3.42/3.79 Clause #11 (by clausification #[10]): ∀ (a : nat), Eq (Ne x (pl y (skS.0 0 a))) False
% 3.42/3.79 Clause #12 (by clausification #[11]): ∀ (a : nat), Eq x (pl y (skS.0 0 a))
% 3.42/3.79 Clause #13 (by clausification #[2]): ∀ (a : nat), Eq (∀ (Xy : nat), Eq (pl a Xy) (pl Xy a)) True
% 3.42/3.79 Clause #14 (by clausification #[13]): ∀ (a a_1 : nat), Eq (Eq (pl a a_1) (pl a_1 a)) True
% 3.42/3.79 Clause #15 (by clausification #[14]): ∀ (a a_1 : nat), Eq (pl a a_1) (pl a_1 a)
% 3.42/3.79 Clause #16 (by clausification #[4]): Eq (Not (∀ (Xx_0 : nat), Ne (pl x z) (pl (pl y z) Xx_0))) False
% 3.42/3.79 Clause #17 (by clausification #[16]): Eq (∀ (Xx_0 : nat), Ne (pl x z) (pl (pl y z) Xx_0)) True
% 3.42/3.79 Clause #18 (by clausification #[17]): ∀ (a : nat), Eq (Ne (pl x z) (pl (pl y z) a)) True
% 3.42/3.79 Clause #19 (by clausification #[18]): ∀ (a : nat), Ne (pl x z) (pl (pl y z) a)
% 3.42/3.79 Clause #20 (by forward demodulation #[19, 15]): ∀ (a : nat), Ne (pl z x) (pl (pl y z) a)
% 3.42/3.79 Clause #21 (by forward demodulation #[20, 15]): ∀ (a : nat), Ne (pl z x) (pl (pl z y) a)
% 3.42/3.79 Clause #23 (by clausification #[3]): ∀ (a : nat), Eq (∀ (Xy Xz : nat), Eq (pl (pl a Xy) Xz) (pl a (pl Xy Xz))) True
% 3.42/3.79 Clause #24 (by clausification #[23]): ∀ (a a_1 : nat), Eq (∀ (Xz : nat), Eq (pl (pl a a_1) Xz) (pl a (pl a_1 Xz))) True
% 3.42/3.79 Clause #25 (by clausification #[24]): ∀ (a a_1 a_2 : nat), Eq (Eq (pl (pl a a_1) a_2) (pl a (pl a_1 a_2))) True
% 3.42/3.79 Clause #26 (by clausification #[25]): ∀ (a a_1 a_2 : nat), Eq (pl (pl a a_1) a_2) (pl a (pl a_1 a_2))
% 3.42/3.79 Clause #27 (by backward demodulation #[26, 21]): ∀ (a : nat), Ne (pl z x) (pl z (pl y a))
% 3.42/3.79 Clause #34 (by superposition #[27, 12]): Ne (pl z x) (pl z x)
% 3.42/3.79 Clause #36 (by eliminate resolved literals #[34]): False
% 3.42/3.79 SZS output end Proof for theBenchmark.p
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