%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : NUM695^1 : 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 10:57:07 EDT 2023

% Result   : Theorem 3.66s 3.89s
% Output   : Proof 3.66s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13  % Problem    : NUM695^1 : TPTP v8.1.2. Released v3.7.0.
% 0.00/0.14  % Command    : duper %s
% 0.19/0.35  % Computer : n026.cluster.edu
% 0.19/0.35  % Model    : x86_64 x86_64
% 0.19/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.19/0.35  % Memory   : 8042.1875MB
% 0.19/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.19/0.35  % CPULimit   : 300
% 0.19/0.35  % WCLimit    : 300
% 0.19/0.35  % DateTime   : Fri Aug 25 17:29:02 EDT 2023
% 0.19/0.35  % CPUTime    : 
% 3.66/3.89  SZS status Theorem for theBenchmark.p
% 3.66/3.89  SZS output start Proof for theBenchmark.p
% 3.66/3.89  Clause #3 (by assumption #[]): Eq (∀ (Xx Xy : nat), more (pl Xx Xy) Xx) True
% 3.66/3.89  Clause #4 (by assumption #[]): Eq (∀ (Xx : nat), Eq (suc Xx) (pl n_1 Xx)) True
% 3.66/3.89  Clause #5 (by assumption #[]): Eq (Not (more (suc x) n_1)) True
% 3.66/3.89  Clause #10 (by clausification #[5]): Eq (more (suc x) n_1) False
% 3.66/3.89  Clause #13 (by clausification #[3]): ∀ (a : nat), Eq (∀ (Xy : nat), more (pl a Xy) a) True
% 3.66/3.89  Clause #14 (by clausification #[13]): ∀ (a a_1 : nat), Eq (more (pl a a_1) a) True
% 3.66/3.89  Clause #15 (by clausification #[4]): ∀ (a : nat), Eq (Eq (suc a) (pl n_1 a)) True
% 3.66/3.89  Clause #16 (by clausification #[15]): ∀ (a : nat), Eq (suc a) (pl n_1 a)
% 3.66/3.89  Clause #17 (by superposition #[16, 14]): ∀ (a : nat), Eq (more (suc a) n_1) True
% 3.66/3.89  Clause #30 (by superposition #[17, 10]): Eq True False
% 3.66/3.89  Clause #33 (by clausification #[30]): False
% 3.66/3.89  SZS output end Proof for theBenchmark.p
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