%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SET590^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n023.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 14:46:56 EDT 2023

% Result   : Theorem 3.49s 3.66s
% Output   : Proof 3.49s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12  % Problem    : SET590^5 : TPTP v8.1.2. Released v4.0.0.
% 0.03/0.13  % Command    : duper %s
% 0.12/0.34  % Computer : n023.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34  % CPULimit   : 300
% 0.12/0.34  % WCLimit    : 300
% 0.12/0.34  % DateTime   : Sat Aug 26 11:33:11 EDT 2023
% 0.12/0.34  % CPUTime    : 
% 3.49/3.66  SZS status Theorem for theBenchmark.p
% 3.49/3.66  SZS output start Proof for theBenchmark.p
% 3.49/3.66  Clause #0 (by assumption #[]): Eq (Not (∀ (X Y : a → Prop) (Xx : a), And (X Xx) (Not (Y Xx)) → X Xx)) True
% 3.49/3.66  Clause #1 (by clausification #[0]): Eq (∀ (X Y : a → Prop) (Xx : a), And (X Xx) (Not (Y Xx)) → X Xx) False
% 3.49/3.66  Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop), Eq (Not (∀ (Y : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (Not (Y Xx)) → skS.0 0 a_1 Xx)) True
% 3.49/3.66  Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop), Eq (∀ (Y : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (Not (Y Xx)) → skS.0 0 a_1 Xx) False
% 3.49/3.66  Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop), Eq (Not (∀ (Xx : a), And (skS.0 0 a_1 Xx) (Not (skS.0 1 a_1 a_2 Xx)) → skS.0 0 a_1 Xx)) True
% 3.49/3.66  Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop), Eq (∀ (Xx : a), And (skS.0 0 a_1 Xx) (Not (skS.0 1 a_1 a_2 Xx)) → skS.0 0 a_1 Xx) False
% 3.49/3.66  Clause #6 (by clausification #[5]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.49/3.66    Eq
% 3.49/3.66      (Not
% 3.49/3.66        (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (Not (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) →
% 3.49/3.66          skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)))
% 3.49/3.66      True
% 3.49/3.66  Clause #7 (by clausification #[6]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.49/3.66    Eq
% 3.49/3.66      (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (Not (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) →
% 3.49/3.66        skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3))
% 3.49/3.66      False
% 3.49/3.66  Clause #8 (by clausification #[7]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.49/3.66    Eq (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (Not (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)))) True
% 3.49/3.66  Clause #9 (by clausification #[7]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) False
% 3.49/3.66  Clause #11 (by clausification #[8]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) True
% 3.49/3.66  Clause #13 (by superposition #[11, 9]): Eq True False
% 3.49/3.66  Clause #14 (by clausification #[13]): False
% 3.49/3.66  SZS output end Proof for theBenchmark.p
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