TSTP Solution File: NUM713^1 by Duper---1.0
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% File : Duper---1.0
% Problem : NUM713^1 : TPTP v8.1.2. Released v3.7.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n031.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 10:57:19 EDT 2023
% Result : Theorem 3.56s 3.71s
% Output : Proof 3.56s
% 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 : NUM713^1 : TPTP v8.1.2. Released v3.7.0.
% 0.00/0.13 % Command : duper %s
% 0.14/0.34 % Computer : n031.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Fri Aug 25 09:19:37 EDT 2023
% 0.14/0.34 % CPUTime :
% 3.56/3.71 SZS status Theorem for theBenchmark.p
% 3.56/3.71 SZS output start Proof for theBenchmark.p
% 3.56/3.71 Clause #0 (by assumption #[]): Eq (Not (∀ (Xx_0 : nat), Ne x (pl y Xx_0))) True
% 3.56/3.71 Clause #2 (by assumption #[]): Eq (∀ (Xx Xy : nat), Eq (ts Xx Xy) (ts Xy Xx)) True
% 3.56/3.71 Clause #3 (by assumption #[]): Eq (∀ (Xx Xy Xz : nat), Eq (ts Xx (pl Xy Xz)) (pl (ts Xx Xy) (ts Xx Xz))) True
% 3.56/3.71 Clause #4 (by assumption #[]): Eq (Not (Not (∀ (Xx_0 : nat), Ne (ts x z) (pl (ts y z) Xx_0)))) True
% 3.56/3.71 Clause #9 (by clausification #[0]): Eq (∀ (Xx_0 : nat), Ne x (pl y Xx_0)) False
% 3.56/3.71 Clause #10 (by clausification #[9]): ∀ (a : nat), Eq (Not (Ne x (pl y (skS.0 0 a)))) True
% 3.56/3.71 Clause #11 (by clausification #[10]): ∀ (a : nat), Eq (Ne x (pl y (skS.0 0 a))) False
% 3.56/3.71 Clause #12 (by clausification #[11]): ∀ (a : nat), Eq x (pl y (skS.0 0 a))
% 3.56/3.71 Clause #13 (by clausification #[2]): ∀ (a : nat), Eq (∀ (Xy : nat), Eq (ts a Xy) (ts Xy a)) True
% 3.56/3.71 Clause #14 (by clausification #[13]): ∀ (a a_1 : nat), Eq (Eq (ts a a_1) (ts a_1 a)) True
% 3.56/3.71 Clause #15 (by clausification #[14]): ∀ (a a_1 : nat), Eq (ts a a_1) (ts a_1 a)
% 3.56/3.72 Clause #16 (by clausification #[4]): Eq (Not (∀ (Xx_0 : nat), Ne (ts x z) (pl (ts y z) Xx_0))) False
% 3.56/3.72 Clause #17 (by clausification #[16]): Eq (∀ (Xx_0 : nat), Ne (ts x z) (pl (ts y z) Xx_0)) True
% 3.56/3.72 Clause #18 (by clausification #[17]): ∀ (a : nat), Eq (Ne (ts x z) (pl (ts y z) a)) True
% 3.56/3.72 Clause #19 (by clausification #[18]): ∀ (a : nat), Ne (ts x z) (pl (ts y z) a)
% 3.56/3.72 Clause #20 (by forward demodulation #[19, 15]): ∀ (a : nat), Ne (ts z x) (pl (ts y z) a)
% 3.56/3.72 Clause #21 (by forward demodulation #[20, 15]): ∀ (a : nat), Ne (ts z x) (pl (ts z y) a)
% 3.56/3.72 Clause #22 (by clausification #[3]): ∀ (a : nat), Eq (∀ (Xy Xz : nat), Eq (ts a (pl Xy Xz)) (pl (ts a Xy) (ts a Xz))) True
% 3.56/3.72 Clause #23 (by clausification #[22]): ∀ (a a_1 : nat), Eq (∀ (Xz : nat), Eq (ts a (pl a_1 Xz)) (pl (ts a a_1) (ts a Xz))) True
% 3.56/3.72 Clause #24 (by clausification #[23]): ∀ (a a_1 a_2 : nat), Eq (Eq (ts a (pl a_1 a_2)) (pl (ts a a_1) (ts a a_2))) True
% 3.56/3.72 Clause #25 (by clausification #[24]): ∀ (a a_1 a_2 : nat), Eq (ts a (pl a_1 a_2)) (pl (ts a a_1) (ts a a_2))
% 3.56/3.72 Clause #26 (by superposition #[25, 21]): ∀ (a : nat), Ne (ts z x) (ts z (pl y a))
% 3.56/3.72 Clause #29 (by superposition #[26, 12]): Ne (ts z x) (ts z x)
% 3.56/3.72 Clause #30 (by eliminate resolved literals #[29]): False
% 3.56/3.72 SZS output end Proof for theBenchmark.p
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