TSTP Solution File: NUM693^1 by Duper---1.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% 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 : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----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
%------------------------------------------------------------------------------