%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : NUM640^1 : TPTP v8.1.2. Released v3.7.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n016.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 10:56:40 EDT 2023

% Result   : Theorem 3.52s 3.68s
% Output   : Proof 3.52s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem    : NUM640^1 : TPTP v8.1.2. Released v3.7.0.
% 0.00/0.13  % Command    : duper %s
% 0.14/0.34  % Computer : n016.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
% 0.14/0.35  % DateTime   : Fri Aug 25 18:03:56 EDT 2023
% 0.14/0.35  % CPUTime    : 
% 3.52/3.68  SZS status Theorem for theBenchmark.p
% 3.52/3.68  SZS output start Proof for theBenchmark.p
% 3.52/3.68  Clause #0 (by assumption #[]): Eq (∀ (Xx Xy : nat), Eq (pl Xx (suc Xy)) (suc (pl Xx Xy))) True
% 3.52/3.68  Clause #1 (by assumption #[]): Eq (Not (Eq (suc (pl x y)) (pl x (suc y)))) True
% 3.52/3.68  Clause #2 (by clausification #[1]): Eq (Eq (suc (pl x y)) (pl x (suc y))) False
% 3.52/3.68  Clause #3 (by clausification #[2]): Ne (suc (pl x y)) (pl x (suc y))
% 3.52/3.68  Clause #4 (by clausification #[0]): ∀ (a : nat), Eq (∀ (Xy : nat), Eq (pl a (suc Xy)) (suc (pl a Xy))) True
% 3.52/3.68  Clause #5 (by clausification #[4]): ∀ (a a_1 : nat), Eq (Eq (pl a (suc a_1)) (suc (pl a a_1))) True
% 3.52/3.68  Clause #6 (by clausification #[5]): ∀ (a a_1 : nat), Eq (pl a (suc a_1)) (suc (pl a a_1))
% 3.52/3.68  Clause #8 (by backward contextual literal cutting #[6, 3]): False
% 3.52/3.68  SZS output end Proof for theBenchmark.p
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