%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SEV159^5 : TPTP v8.1.2. Released v4.0.0.
% 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 19:24:22 EDT 2023

% Result   : Theorem 3.37s 3.53s
% Output   : Proof 3.37s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.13  % Problem    : SEV159^5 : TPTP v8.1.2. Released v4.0.0.
% 0.07/0.14  % Command    : duper %s
% 0.14/0.35  % Computer : n031.cluster.edu
% 0.14/0.35  % Model    : x86_64 x86_64
% 0.14/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35  % Memory   : 8042.1875MB
% 0.14/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35  % CPULimit   : 300
% 0.14/0.35  % WCLimit    : 300
% 0.14/0.35  % DateTime   : Thu Aug 24 04:01:51 EDT 2023
% 0.14/0.35  % CPUTime    : 
% 3.37/3.53  SZS status Theorem for theBenchmark.p
% 3.37/3.53  SZS output start Proof for theBenchmark.p
% 3.37/3.53  Clause #0 (by assumption #[]): Eq (Not (a → ∀ (Xy : a), Eq Xy Xy)) True
% 3.37/3.53  Clause #1 (by clausification #[0]): Eq (a → ∀ (Xy : a), Eq Xy Xy) False
% 3.37/3.53  Clause #2 (by clausification #[1]): a → Eq (Not (∀ (Xy : a), Eq Xy Xy)) True
% 3.37/3.53  Clause #3 (by clausification #[2]): Eq (∀ (Xy : a), Eq Xy Xy) False
% 3.37/3.53  Clause #4 (by clausification #[3]): ∀ (a_1 : a), Eq (Not (Eq (skS.0 1 a_1) (skS.0 1 a_1))) True
% 3.37/3.53  Clause #5 (by clausification #[4]): ∀ (a_1 : a), Eq (Eq (skS.0 1 a_1) (skS.0 1 a_1)) False
% 3.37/3.53  Clause #6 (by clausification #[5]): ∀ (a_1 : a), Ne (skS.0 1 a_1) (skS.0 1 a_1)
% 3.37/3.53  Clause #7 (by eliminate resolved literals #[6]): False
% 3.37/3.53  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------