0.05/0.13	% Problem    : theBenchmark.p : TPTP v0.0.0. Released v0.0.0.
0.05/0.14	% Command    : duper %s
0.10/0.35	% Computer : n004.cluster.edu
0.10/0.35	% Model    : x86_64 x86_64
0.10/0.35	% CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
0.10/0.35	% Memory   : 8042.1875MB
0.10/0.35	% OS       : Linux 3.10.0-693.el7.x86_64
0.10/0.35	% CPULimit   : 1440
0.10/0.35	% WCLimit    : 180
0.10/0.35	% DateTime   : Mon Jul  3 04:48:45 EDT 2023
0.10/0.35	% CPUTime    : 
5.55/5.98	SZS status Theorem for theBenchmark.p
5.55/5.98	SZS output start Proof for theBenchmark.p
5.55/5.98	Clause #0 (by assumption #[]): Eq (Not (Not (∀ (Xx_0 : nat), Ne z (pl x Xx_0)))) True
5.55/5.98	Clause #1 (by assumption #[]): Eq (Not (∀ (Xx_0 : nat), Ne y (pl x Xx_0))) True
5.55/5.98	Clause #3 (by assumption #[]): Eq (∀ (Xx Xy Xz : nat), Eq (pl (pl Xx Xy) Xz) (pl Xx (pl Xy Xz))) True
5.55/5.98	Clause #4 (by assumption #[]): Eq (Not (∀ (Xx : nat), Ne z (pl y Xx))) True
5.55/5.98	Clause #9 (by clausification #[1]): Eq (∀ (Xx_0 : nat), Ne y (pl x Xx_0)) False
5.55/5.98	Clause #10 (by clausification #[9]): ∀ (a : nat), Eq (Not (Ne y (pl x (skS.0 0 a)))) True
5.55/5.98	Clause #11 (by clausification #[10]): ∀ (a : nat), Eq (Ne y (pl x (skS.0 0 a))) False
5.55/5.98	Clause #12 (by clausification #[11]): ∀ (a : nat), Eq y (pl x (skS.0 0 a))
5.55/5.98	Clause #13 (by clausification #[4]): Eq (∀ (Xx : nat), Ne z (pl y Xx)) False
5.55/5.98	Clause #14 (by clausification #[13]): ∀ (a : nat), Eq (Not (Ne z (pl y (skS.0 1 a)))) True
5.55/5.98	Clause #15 (by clausification #[14]): ∀ (a : nat), Eq (Ne z (pl y (skS.0 1 a))) False
5.55/5.98	Clause #16 (by clausification #[15]): ∀ (a : nat), Eq z (pl y (skS.0 1 a))
5.55/5.98	Clause #17 (by clausification #[0]): Eq (Not (∀ (Xx_0 : nat), Ne z (pl x Xx_0))) False
5.55/5.98	Clause #18 (by clausification #[17]): Eq (∀ (Xx_0 : nat), Ne z (pl x Xx_0)) True
5.55/5.98	Clause #19 (by clausification #[18]): ∀ (a : nat), Eq (Ne z (pl x a)) True
5.55/5.98	Clause #20 (by clausification #[19]): ∀ (a : nat), Ne z (pl x a)
5.55/5.98	Clause #22 (by clausification #[3]): ∀ (a : nat), Eq (∀ (Xy Xz : nat), Eq (pl (pl a Xy) Xz) (pl a (pl Xy Xz))) True
5.55/5.98	Clause #23 (by clausification #[22]): ∀ (a a_1 : nat), Eq (∀ (Xz : nat), Eq (pl (pl a a_1) Xz) (pl a (pl a_1 Xz))) True
5.55/5.98	Clause #24 (by clausification #[23]): ∀ (a a_1 a_2 : nat), Eq (Eq (pl (pl a a_1) a_2) (pl a (pl a_1 a_2))) True
5.55/5.98	Clause #25 (by clausification #[24]): ∀ (a a_1 a_2 : nat), Eq (pl (pl a a_1) a_2) (pl a (pl a_1 a_2))
5.55/5.98	Clause #27 (by superposition #[25, 12]): ∀ (a a_1 : nat), Eq (pl y a) (pl x (pl (skS.0 0 a_1) a))
5.55/5.98	Clause #30 (by superposition #[27, 20]): ∀ (a : nat), Ne z (pl y a)
5.55/5.98	Clause #32 (by backward contextual literal cutting #[30, 16]): False
5.55/5.98	SZS output end Proof for theBenchmark.p
5.55/5.98	EOF