%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SYO142^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n027.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 : Fri Sep  1 04:21:38 EDT 2023

% Result   : Theorem 3.89s 4.07s
% Output   : Proof 3.89s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.15  % Problem    : SYO142^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.16  % Command    : duper %s
% 0.16/0.38  % Computer : n027.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   : Sat Aug 26 05:43:19 EDT 2023
% 0.16/0.38  % CPUTime    : 
% 3.89/4.07  SZS status Theorem for theBenchmark.p
% 3.89/4.07  SZS output start Proof for theBenchmark.p
% 3.89/4.07  Clause #0 (by assumption #[]): Eq (Not (Exists fun Xx => Exists fun Xy => And (p Xx → p (f (f Xy))) (p Xy → p (f (f a))))) True
% 3.89/4.07  Clause #1 (by clausification #[0]): Eq (Exists fun Xx => Exists fun Xy => And (p Xx → p (f (f Xy))) (p Xy → p (f (f a)))) False
% 3.89/4.07  Clause #2 (by clausification #[1]): ∀ (a_1 : Iota), Eq (Exists fun Xy => And (p a_1 → p (f (f Xy))) (p Xy → p (f (f a)))) False
% 3.89/4.07  Clause #3 (by clausification #[2]): ∀ (a_1 a_2 : Iota), Eq (And (p a_1 → p (f (f a_2))) (p a_2 → p (f (f a)))) False
% 3.89/4.07  Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : Iota), Or (Eq (p a_1 → p (f (f a_2))) False) (Eq (p a_2 → p (f (f a))) False)
% 3.89/4.07  Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : Iota), Or (Eq (p a_1 → p (f (f a))) False) (Eq (p a_2) True)
% 3.89/4.07  Clause #6 (by clausification #[4]): ∀ (a_1 : Iota), Or (Eq (p a_1 → p (f (f a))) False) (Eq (p (f (f a_1))) False)
% 3.89/4.07  Clause #7 (by clausification #[5]): ∀ (a a_1 : Iota), Or (Eq (p a) True) (Eq (p a_1) True)
% 3.89/4.07  Clause #9 (by equality factoring #[7]): ∀ (a : Iota), Or (Ne True True) (Eq (p a) True)
% 3.89/4.07  Clause #10 (by clausification #[9]): ∀ (a : Iota), Or (Eq (p a) True) (Or (Eq True False) (Eq True False))
% 3.89/4.07  Clause #12 (by clausification #[10]): ∀ (a : Iota), Or (Eq (p a) True) (Eq True False)
% 3.89/4.07  Clause #13 (by clausification #[12]): ∀ (a : Iota), Eq (p a) True
% 3.89/4.07  Clause #15 (by clausification #[6]): ∀ (a_1 : Iota), Or (Eq (p (f (f a_1))) False) (Eq (p (f (f a))) False)
% 3.89/4.07  Clause #17 (by superposition #[15, 13]): Or (Eq (p (f (f a))) False) (Eq False True)
% 3.89/4.07  Clause #18 (by clausification #[17]): Eq (p (f (f a))) False
% 3.89/4.07  Clause #19 (by superposition #[18, 13]): Eq False True
% 3.89/4.07  Clause #20 (by clausification #[19]): False
% 3.89/4.07  SZS output end Proof for theBenchmark.p
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