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Duper---1.0.THM-Prf.s

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%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SET027^5 : TPTP v9.2.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n008.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 07:59:25 PM UTC 2025

% Result   : Theorem 3.63s 3.86s
% Output   : Proof 3.63s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.10/0.12  % Problem    : SET027^5 : TPTP v9.2.0. Released v4.0.0.
% 0.10/0.13  % Command    : duper %s
% 0.13/0.34  % Computer : n008.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit   : 300
% 0.13/0.34  % WCLimit    : 300
% 0.13/0.34  % DateTime   : Thu Oct  2 14:46:53 EDT 2025
% 0.13/0.34  % CPUTime    : 
% 3.63/3.86  SZS status Theorem for theBenchmark.p
% 3.63/3.86  SZS output start Proof for theBenchmark.p
% 3.63/3.86  Clause #0 (by assumption #[]): Eq (Not (∀ (X Y Z : a → Prop), And (∀ (Xx : a), X Xx → Y Xx) (∀ (Xx : a), Y Xx → Z Xx) → ∀ (Xx : a), X Xx → Z Xx)) True
% 3.63/3.86  Clause #1 (by clausification #[0]): Eq (∀ (X Y Z : a → Prop), And (∀ (Xx : a), X Xx → Y Xx) (∀ (Xx : a), Y Xx → Z Xx) → ∀ (Xx : a), X Xx → Z Xx) False
% 3.63/3.86  Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop),
% 3.63/3.86    Eq
% 3.63/3.86      (Not
% 3.63/3.86        (∀ (Y Z : a → Prop),
% 3.63/3.86          And (∀ (Xx : a), skS.0 0 a_1 Xx → Y Xx) (∀ (Xx : a), Y Xx → Z Xx) → ∀ (Xx : a), skS.0 0 a_1 Xx → Z Xx))
% 3.63/3.86      True
% 3.63/3.86  Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop),
% 3.63/3.86    Eq
% 3.63/3.86      (∀ (Y Z : a → Prop),
% 3.63/3.86        And (∀ (Xx : a), skS.0 0 a_1 Xx → Y Xx) (∀ (Xx : a), Y Xx → Z Xx) → ∀ (Xx : a), skS.0 0 a_1 Xx → Z Xx)
% 3.63/3.86      False
% 3.63/3.86  Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 3.63/3.86    Eq
% 3.63/3.86      (Not
% 3.63/3.86        (∀ (Z : a → Prop),
% 3.63/3.86          And (∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 1 a_1 a_2 Xx) (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → Z Xx) →
% 3.63/3.86            ∀ (Xx : a), skS.0 0 a_1 Xx → Z Xx))
% 3.63/3.86      True
% 3.63/3.86  Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 3.63/3.86    Eq
% 3.63/3.86      (∀ (Z : a → Prop),
% 3.63/3.86        And (∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 1 a_1 a_2 Xx) (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → Z Xx) →
% 3.63/3.86          ∀ (Xx : a), skS.0 0 a_1 Xx → Z Xx)
% 3.63/3.86      False
% 3.63/3.86  Clause #6 (by clausification #[5]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.63/3.86    Eq
% 3.63/3.86      (Not
% 3.63/3.86        (And (∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 1 a_1 a_2 Xx) (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → skS.0 2 a_1 a_2 a_3 Xx) →
% 3.63/3.86          ∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 2 a_1 a_2 a_3 Xx))
% 3.63/3.86      True
% 3.63/3.86  Clause #7 (by clausification #[6]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.63/3.86    Eq
% 3.63/3.86      (And (∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 1 a_1 a_2 Xx) (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → skS.0 2 a_1 a_2 a_3 Xx) →
% 3.63/3.86        ∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 2 a_1 a_2 a_3 Xx)
% 3.63/3.86      False
% 3.63/3.86  Clause #8 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.63/3.86    Eq (And (∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 1 a_1 a_2 Xx) (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → skS.0 2 a_1 a_2 a_3 Xx))
% 3.63/3.86      True
% 3.63/3.86  Clause #9 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop), Eq (∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 2 a_1 a_2 a_3 Xx) False
% 3.63/3.86  Clause #10 (by clausification #[8]): ∀ (a_1 a_2 a_3 : a → Prop), Eq (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → skS.0 2 a_1 a_2 a_3 Xx) True
% 3.63/3.86  Clause #11 (by clausification #[8]): ∀ (a_1 a_2 : a → Prop), Eq (∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 1 a_1 a_2 Xx) True
% 3.63/3.86  Clause #12 (by clausification #[10]): ∀ (a_1 a_2 : a → Prop) (a_3 : a) (a_4 : a → Prop), Eq (skS.0 1 a_1 a_2 a_3 → skS.0 2 a_1 a_2 a_4 a_3) True
% 3.63/3.86  Clause #13 (by clausification #[12]): ∀ (a_1 a_2 : a → Prop) (a_3 : a) (a_4 : a → Prop),
% 3.63/3.86    Or (Eq (skS.0 1 a_1 a_2 a_3) False) (Eq (skS.0 2 a_1 a_2 a_4 a_3) True)
% 3.63/3.86  Clause #14 (by clausification #[11]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 : a → Prop), Eq (skS.0 0 a_1 a_2 → skS.0 1 a_1 a_3 a_2) True
% 3.63/3.86  Clause #15 (by clausification #[14]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 : a → Prop), Or (Eq (skS.0 0 a_1 a_2) False) (Eq (skS.0 1 a_1 a_3 a_2) True)
% 3.63/3.86  Clause #16 (by clausification #[9]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a),
% 3.63/3.86    Eq (Not (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4) → skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4))) True
% 3.63/3.86  Clause #17 (by clausification #[16]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a),
% 3.63/3.86    Eq (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4) → skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4)) False
% 3.63/3.86  Clause #18 (by clausification #[17]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) True
% 3.63/3.86  Clause #19 (by clausification #[17]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4)) False
% 3.63/3.86  Clause #20 (by superposition #[18, 15]): ∀ (a_1 a_2 a_3 a_4 : a → Prop) (a_5 : a), Or (Eq True False) (Eq (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_3 a_4 a_5)) True)
% 3.63/3.86  Clause #21 (by clausification #[20]): ∀ (a_1 a_2 a_3 a_4 : a → Prop) (a_5 : a), Eq (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_3 a_4 a_5)) True
% 3.63/3.87  Clause #22 (by superposition #[21, 13]): ∀ (a_1 a_2 a_3 a_4 a_5 : a → Prop) (a_6 : a),
% 3.63/3.87    Or (Eq True False) (Eq (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_4 a_5 a_6)) True)
% 3.63/3.87  Clause #23 (by clausification #[22]): ∀ (a_1 a_2 a_3 a_4 a_5 : a → Prop) (a_6 : a), Eq (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_4 a_5 a_6)) True
% 3.63/3.87  Clause #24 (by superposition #[23, 19]): Eq True False
% 3.63/3.87  Clause #25 (by clausification #[24]): False
% 3.63/3.87  SZS output end Proof for theBenchmark.p
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