TSTP Solution File: SEU936^5 by Duper---1.0

View Problem - Process Solution

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% File     : Duper---1.0
% Problem  : SEU936^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n011.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 16:44:21 EDT 2023

% Result   : Theorem 4.33s 4.49s
% Output   : Proof 4.33s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.07  % Problem    : SEU936^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.07  % Command    : duper %s
% 0.06/0.27  % Computer : n011.cluster.edu
% 0.06/0.27  % Model    : x86_64 x86_64
% 0.06/0.27  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.06/0.27  % Memory   : 8042.1875MB
% 0.06/0.27  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.06/0.27  % CPULimit   : 300
% 0.06/0.27  % WCLimit    : 300
% 0.06/0.27  % DateTime   : Wed Aug 23 14:53:07 EDT 2023
% 0.06/0.27  % CPUTime    : 
% 4.33/4.49  SZS status Theorem for theBenchmark.p
% 4.33/4.49  SZS output start Proof for theBenchmark.p
% 4.33/4.49  Clause #0 (by assumption #[]): Eq
% 4.33/4.49    (Not
% 4.33/4.49      (∀ (F : a → b) (G : b → c),
% 4.33/4.49        (∀ (Xx Xy : a), Eq (G (F Xx)) (G (F Xy)) → Eq Xx Xy) → ∀ (Xx Xy : a), Eq (F Xx) (F Xy) → Eq Xx Xy))
% 4.33/4.49    True
% 4.33/4.49  Clause #1 (by clausification #[0]): Eq
% 4.33/4.49    (∀ (F : a → b) (G : b → c),
% 4.33/4.49      (∀ (Xx Xy : a), Eq (G (F Xx)) (G (F Xy)) → Eq Xx Xy) → ∀ (Xx Xy : a), Eq (F Xx) (F Xy) → Eq Xx Xy)
% 4.33/4.49    False
% 4.33/4.49  Clause #2 (by clausification #[1]): ∀ (a_1 : a → b),
% 4.33/4.49    Eq
% 4.33/4.49      (Not
% 4.33/4.49        (∀ (G : b → c),
% 4.33/4.49          (∀ (Xx Xy : a), Eq (G (skS.0 0 a_1 Xx)) (G (skS.0 0 a_1 Xy)) → Eq Xx Xy) →
% 4.33/4.49            ∀ (Xx Xy : a), Eq (skS.0 0 a_1 Xx) (skS.0 0 a_1 Xy) → Eq Xx Xy))
% 4.33/4.49      True
% 4.33/4.49  Clause #3 (by clausification #[2]): ∀ (a_1 : a → b),
% 4.33/4.49    Eq
% 4.33/4.49      (∀ (G : b → c),
% 4.33/4.49        (∀ (Xx Xy : a), Eq (G (skS.0 0 a_1 Xx)) (G (skS.0 0 a_1 Xy)) → Eq Xx Xy) →
% 4.33/4.49          ∀ (Xx Xy : a), Eq (skS.0 0 a_1 Xx) (skS.0 0 a_1 Xy) → Eq Xx Xy)
% 4.33/4.49      False
% 4.33/4.49  Clause #4 (by clausification #[3]): ∀ (a_1 : a → b) (a_2 : b → c),
% 4.33/4.49    Eq
% 4.33/4.49      (Not
% 4.33/4.49        ((∀ (Xx Xy : a), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xx)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xy)) → Eq Xx Xy) →
% 4.33/4.49          ∀ (Xx Xy : a), Eq (skS.0 0 a_1 Xx) (skS.0 0 a_1 Xy) → Eq Xx Xy))
% 4.33/4.49      True
% 4.33/4.49  Clause #5 (by clausification #[4]): ∀ (a_1 : a → b) (a_2 : b → c),
% 4.33/4.49    Eq
% 4.33/4.49      ((∀ (Xx Xy : a), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xx)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xy)) → Eq Xx Xy) →
% 4.33/4.49        ∀ (Xx Xy : a), Eq (skS.0 0 a_1 Xx) (skS.0 0 a_1 Xy) → Eq Xx Xy)
% 4.33/4.49      False
% 4.33/4.49  Clause #6 (by clausification #[5]): ∀ (a_1 : a → b) (a_2 : b → c),
% 4.33/4.49    Eq (∀ (Xx Xy : a), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xx)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xy)) → Eq Xx Xy) True
% 4.33/4.49  Clause #7 (by clausification #[5]): ∀ (a_1 : a → b), Eq (∀ (Xx Xy : a), Eq (skS.0 0 a_1 Xx) (skS.0 0 a_1 Xy) → Eq Xx Xy) False
% 4.33/4.49  Clause #8 (by clausification #[6]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : a),
% 4.33/4.49    Eq (∀ (Xy : a), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_3)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xy)) → Eq a_3 Xy) True
% 4.33/4.49  Clause #9 (by clausification #[8]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 a_4 : a),
% 4.33/4.49    Eq (Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_3)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_4)) → Eq a_3 a_4) True
% 4.33/4.49  Clause #10 (by clausification #[9]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 a_4 : a),
% 4.33/4.49    Or (Eq (Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_3)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_4))) False) (Eq (Eq a_3 a_4) True)
% 4.33/4.49  Clause #11 (by clausification #[10]): ∀ (a_1 a_2 : a) (a_3 : a → b) (a_4 : b → c),
% 4.33/4.49    Or (Eq (Eq a_1 a_2) True) (Ne (skS.0 1 a_3 a_4 (skS.0 0 a_3 a_1)) (skS.0 1 a_3 a_4 (skS.0 0 a_3 a_2)))
% 4.33/4.49  Clause #12 (by clausification #[11]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 a_4 : a),
% 4.33/4.49    Or (Ne (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_3)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_4))) (Eq a_3 a_4)
% 4.33/4.49  Clause #14 (by clausification #[7]): ∀ (a_1 : a → b) (a_2 : a),
% 4.33/4.49    Eq (Not (∀ (Xy : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 Xy) → Eq (skS.0 2 a_1 a_2) Xy)) True
% 4.33/4.49  Clause #15 (by clausification #[14]): ∀ (a_1 : a → b) (a_2 : a),
% 4.33/4.49    Eq (∀ (Xy : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 Xy) → Eq (skS.0 2 a_1 a_2) Xy) False
% 4.33/4.49  Clause #16 (by clausification #[15]): ∀ (a_1 : a → b) (a_2 a_3 : a),
% 4.33/4.49    Eq
% 4.33/4.49      (Not
% 4.33/4.49        (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3)) →
% 4.33/4.49          Eq (skS.0 2 a_1 a_2) (skS.0 3 a_1 a_2 a_3)))
% 4.33/4.49      True
% 4.33/4.49  Clause #17 (by clausification #[16]): ∀ (a_1 : a → b) (a_2 a_3 : a),
% 4.33/4.49    Eq
% 4.33/4.49      (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3)) →
% 4.33/4.49        Eq (skS.0 2 a_1 a_2) (skS.0 3 a_1 a_2 a_3))
% 4.33/4.49      False
% 4.33/4.49  Clause #18 (by clausification #[17]): ∀ (a_1 : a → b) (a_2 a_3 : a), Eq (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3))) True
% 4.33/4.49  Clause #19 (by clausification #[17]): ∀ (a_1 : a → b) (a_2 a_3 : a), Eq (Eq (skS.0 2 a_1 a_2) (skS.0 3 a_1 a_2 a_3)) False
% 4.33/4.49  Clause #20 (by clausification #[18]): ∀ (a_1 : a → b) (a_2 a_3 : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3))
% 4.33/4.50  Clause #21 (by superposition #[20, 12]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 a_4 a_5 : a),
% 4.33/4.50    Or (Ne (skS.0 1 a_1 a_2 (skS.0 0 a_1 (skS.0 2 a_1 a_3))) (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_4)))
% 4.33/4.50      (Eq (skS.0 3 a_1 a_3 a_5) a_4)
% 4.33/4.50  Clause #23 (by clausification #[19]): ∀ (a_1 : a → b) (a_2 a_3 : a), Ne (skS.0 2 a_1 a_2) (skS.0 3 a_1 a_2 a_3)
% 4.33/4.50  Clause #24 (by equality resolution #[21]): ∀ (a_1 : a → b) (a_2 a_3 : a), Eq (skS.0 3 a_1 a_2 a_3) (skS.0 2 a_1 a_2)
% 4.33/4.50  Clause #26 (by forward contextual literal cutting #[24, 23]): False
% 4.33/4.50  SZS output end Proof for theBenchmark.p
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