TSTP Solution File: NUM693^1 by Duper---1.0
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% File : Duper---1.0
% Problem : NUM693^1 : TPTP v8.1.2. Released v3.7.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n026.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 10:57:05 EDT 2023
% Result : Theorem 3.50s 3.76s
% Output : Proof 3.50s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.15 % Problem : NUM693^1 : TPTP v8.1.2. Released v3.7.0.
% 0.00/0.16 % Command : duper %s
% 0.16/0.38 % Computer : n026.cluster.edu
% 0.16/0.38 % Model : x86_64 x86_64
% 0.16/0.38 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.16/0.38 % Memory : 8042.1875MB
% 0.16/0.38 % OS : Linux 3.10.0-693.el7.x86_64
% 0.16/0.38 % CPULimit : 300
% 0.16/0.38 % WCLimit : 300
% 0.16/0.38 % DateTime : Fri Aug 25 18:14:48 EDT 2023
% 0.16/0.38 % CPUTime :
% 3.50/3.76 SZS status Theorem for theBenchmark.p
% 3.50/3.76 SZS output start Proof for theBenchmark.p
% 3.50/3.76 Clause #1 (by assumption #[]): Eq (∀ (Xx : nat), Ne Xx n_1 → Not (∀ (Xx_0 : nat), Ne Xx (suc Xx_0))) True
% 3.50/3.76 Clause #2 (by assumption #[]): Eq (∀ (Xx Xy : nat), more (pl Xx Xy) Xx) True
% 3.50/3.76 Clause #3 (by assumption #[]): Eq (∀ (Xx : nat), Eq (suc Xx) (pl n_1 Xx)) True
% 3.50/3.76 Clause #4 (by assumption #[]): Eq (Not (Not (more x n_1) → Eq x n_1)) True
% 3.50/3.76 Clause #9 (by clausification #[4]): Eq (Not (more x n_1) → Eq x n_1) False
% 3.50/3.76 Clause #10 (by clausification #[9]): Eq (Not (more x n_1)) True
% 3.50/3.76 Clause #11 (by clausification #[9]): Eq (Eq x n_1) False
% 3.50/3.76 Clause #12 (by clausification #[10]): Eq (more x n_1) False
% 3.50/3.76 Clause #13 (by clausification #[11]): Ne x n_1
% 3.50/3.76 Clause #14 (by clausification #[2]): ∀ (a : nat), Eq (∀ (Xy : nat), more (pl a Xy) a) True
% 3.50/3.76 Clause #15 (by clausification #[14]): ∀ (a a_1 : nat), Eq (more (pl a a_1) a) True
% 3.50/3.76 Clause #16 (by clausification #[3]): ∀ (a : nat), Eq (Eq (suc a) (pl n_1 a)) True
% 3.50/3.76 Clause #17 (by clausification #[16]): ∀ (a : nat), Eq (suc a) (pl n_1 a)
% 3.50/3.76 Clause #18 (by superposition #[17, 15]): ∀ (a : nat), Eq (more (suc a) n_1) True
% 3.50/3.76 Clause #19 (by clausification #[1]): ∀ (a : nat), Eq (Ne a n_1 → Not (∀ (Xx_0 : nat), Ne a (suc Xx_0))) True
% 3.50/3.76 Clause #20 (by clausification #[19]): ∀ (a : nat), Or (Eq (Ne a n_1) False) (Eq (Not (∀ (Xx_0 : nat), Ne a (suc Xx_0))) True)
% 3.50/3.76 Clause #21 (by clausification #[20]): ∀ (a : nat), Or (Eq (Not (∀ (Xx_0 : nat), Ne a (suc Xx_0))) True) (Eq a n_1)
% 3.50/3.76 Clause #22 (by clausification #[21]): ∀ (a : nat), Or (Eq a n_1) (Eq (∀ (Xx_0 : nat), Ne a (suc Xx_0)) False)
% 3.50/3.76 Clause #23 (by clausification #[22]): ∀ (a a_1 : nat), Or (Eq a n_1) (Eq (Not (Ne a (suc (skS.0 0 a a_1)))) True)
% 3.50/3.76 Clause #24 (by clausification #[23]): ∀ (a a_1 : nat), Or (Eq a n_1) (Eq (Ne a (suc (skS.0 0 a a_1))) False)
% 3.50/3.76 Clause #25 (by clausification #[24]): ∀ (a a_1 : nat), Or (Eq a n_1) (Eq a (suc (skS.0 0 a a_1)))
% 3.50/3.76 Clause #31 (by superposition #[18, 25]): ∀ (a : nat), Or (Eq a n_1) (Eq (more a n_1) True)
% 3.50/3.76 Clause #37 (by superposition #[31, 12]): Or (Eq x n_1) (Eq True False)
% 3.50/3.76 Clause #39 (by clausification #[37]): Eq x n_1
% 3.50/3.76 Clause #40 (by forward contextual literal cutting #[39, 13]): False
% 3.50/3.76 SZS output end Proof for theBenchmark.p
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