TSTP Solution File: NUM703^1 by Duper---1.0
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% File : Duper---1.0
% Problem : NUM703^1 : TPTP v8.1.2. Released v3.7.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n013.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 10:57:15 EDT 2023
% Result : Theorem 3.27s 3.59s
% Output : Proof 3.27s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : NUM703^1 : TPTP v8.1.2. Released v3.7.0.
% 0.00/0.14 % Command : duper %s
% 0.14/0.35 % Computer : n013.cluster.edu
% 0.14/0.35 % Model : x86_64 x86_64
% 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35 % Memory : 8042.1875MB
% 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35 % CPULimit : 300
% 0.14/0.35 % WCLimit : 300
% 0.14/0.35 % DateTime : Fri Aug 25 14:27:17 EDT 2023
% 0.14/0.35 % CPUTime :
% 3.27/3.59 SZS status Theorem for theBenchmark.p
% 3.27/3.59 SZS output start Proof for theBenchmark.p
% 3.27/3.59 Clause #0 (by assumption #[]): Eq (more (suc y) x) True
% 3.27/3.59 Clause #1 (by assumption #[]): Eq (∀ (Xx Xy : nat), more (pl Xy n_1) Xx → moreis Xy Xx) True
% 3.27/3.59 Clause #2 (by assumption #[]): Eq (∀ (Xx : nat), Eq (suc Xx) (pl Xx n_1)) True
% 3.27/3.59 Clause #3 (by assumption #[]): Eq (Not (moreis y x)) True
% 3.27/3.59 Clause #4 (by clausification #[3]): Eq (moreis y x) False
% 3.27/3.59 Clause #5 (by clausification #[1]): ∀ (a : nat), Eq (∀ (Xy : nat), more (pl Xy n_1) a → moreis Xy a) True
% 3.27/3.59 Clause #6 (by clausification #[5]): ∀ (a a_1 : nat), Eq (more (pl a n_1) a_1 → moreis a a_1) True
% 3.27/3.59 Clause #7 (by clausification #[6]): ∀ (a a_1 : nat), Or (Eq (more (pl a n_1) a_1) False) (Eq (moreis a a_1) True)
% 3.27/3.59 Clause #8 (by clausification #[2]): ∀ (a : nat), Eq (Eq (suc a) (pl a n_1)) True
% 3.27/3.59 Clause #9 (by clausification #[8]): ∀ (a : nat), Eq (suc a) (pl a n_1)
% 3.27/3.59 Clause #10 (by backward demodulation #[9, 7]): ∀ (a a_1 : nat), Or (Eq (more (suc a) a_1) False) (Eq (moreis a a_1) True)
% 3.27/3.59 Clause #11 (by superposition #[10, 0]): Or (Eq (moreis y x) True) (Eq False True)
% 3.27/3.59 Clause #12 (by clausification #[11]): Eq (moreis y x) True
% 3.27/3.59 Clause #13 (by superposition #[12, 4]): Eq True False
% 3.27/3.59 Clause #14 (by clausification #[13]): False
% 3.27/3.59 SZS output end Proof for theBenchmark.p
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