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Duper---1.0.THM-Prf.s

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%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SET634^5 : TPTP v9.2.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n010.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 Oct  3 08:00:19 PM UTC 2025

% Result   : Theorem 5.22s 5.44s
% Output   : Proof 5.22s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.10/0.11  % Problem    : SET634^5 : TPTP v9.2.0. Released v4.0.0.
% 0.10/0.12  % Command    : duper %s
% 0.12/0.33  % Computer : n010.cluster.edu
% 0.12/0.33  % Model    : x86_64 x86_64
% 0.12/0.33  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33  % Memory   : 8042.1875MB
% 0.12/0.33  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33  % CPULimit   : 300
% 0.12/0.33  % WCLimit    : 300
% 0.12/0.33  % DateTime   : Thu Oct  2 14:31:53 EDT 2025
% 0.12/0.33  % CPUTime    : 
% 5.22/5.44  SZS status Theorem for theBenchmark.p
% 5.22/5.44  SZS output start Proof for theBenchmark.p
% 5.22/5.44  Clause #0 (by assumption #[]): Eq
% 5.22/5.44    (Not
% 5.22/5.44      (∀ (X Y Z : a → Prop),
% 5.22/5.44        Eq (fun Xx => And (And (X Xx) (Y Xx)) (Not (Z Xx))) fun Xx => And (And (X Xx) (Y Xx)) (Not (Z Xx))))
% 5.22/5.44    True
% 5.22/5.44  Clause #1 (by clausification #[0]): Eq
% 5.22/5.44    (∀ (X Y Z : a → Prop),
% 5.22/5.44      Eq (fun Xx => And (And (X Xx) (Y Xx)) (Not (Z Xx))) fun Xx => And (And (X Xx) (Y Xx)) (Not (Z Xx)))
% 5.22/5.44    False
% 5.22/5.44  Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop),
% 5.22/5.44    Eq
% 5.22/5.44      (Not
% 5.22/5.44        (∀ (Y Z : a → Prop),
% 5.22/5.44          Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (Not (Z Xx))) fun Xx =>
% 5.22/5.44            And (And (skS.0 0 a_1 Xx) (Y Xx)) (Not (Z Xx))))
% 5.22/5.44      True
% 5.22/5.44  Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop),
% 5.22/5.44    Eq
% 5.22/5.44      (∀ (Y Z : a → Prop),
% 5.22/5.44        Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (Not (Z Xx))) fun Xx =>
% 5.22/5.44          And (And (skS.0 0 a_1 Xx) (Y Xx)) (Not (Z Xx)))
% 5.22/5.44      False
% 5.22/5.44  Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 5.22/5.44    Eq
% 5.22/5.44      (Not
% 5.22/5.44        (∀ (Z : a → Prop),
% 5.22/5.44          Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (Z Xx))) fun Xx =>
% 5.22/5.44            And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (Z Xx))))
% 5.22/5.44      True
% 5.22/5.44  Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 5.22/5.44    Eq
% 5.22/5.44      (∀ (Z : a → Prop),
% 5.22/5.44        Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (Z Xx))) fun Xx =>
% 5.22/5.44          And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (Z Xx)))
% 5.22/5.44      False
% 5.22/5.44  Clause #6 (by clausification #[5]): ∀ (a_1 a_2 a_3 : a → Prop),
% 5.22/5.44    Eq
% 5.22/5.44      (Not
% 5.22/5.44        (Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))) fun Xx =>
% 5.22/5.44          And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))))
% 5.22/5.44      True
% 5.22/5.44  Clause #7 (by clausification #[6]): ∀ (a_1 a_2 a_3 : a → Prop),
% 5.22/5.44    Eq
% 5.22/5.44      (Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))) fun Xx =>
% 5.22/5.44        And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx)))
% 5.22/5.44      False
% 5.22/5.44  Clause #8 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop),
% 5.22/5.44    Ne (fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))) fun Xx =>
% 5.22/5.44      And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))
% 5.22/5.44  Clause #9 (by eliminate resolved literals #[8]): False
% 5.22/5.44  SZS output end Proof for theBenchmark.p
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