TSTP Solution File: SEU927^5 by Duper---1.0

View Problem - Process Solution

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% File     : Duper---1.0
% Problem  : SEU927^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n029.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 16:44:19 EDT 2023

% Result   : Theorem 3.58s 3.85s
% Output   : Proof 3.58s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13  % Problem    : SEU927^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14  % Command    : duper %s
% 0.14/0.36  % Computer : n029.cluster.edu
% 0.14/0.36  % Model    : x86_64 x86_64
% 0.14/0.36  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.36  % Memory   : 8042.1875MB
% 0.14/0.36  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.36  % CPULimit   : 300
% 0.14/0.36  % WCLimit    : 300
% 0.14/0.36  % DateTime   : Wed Aug 23 16:12:08 EDT 2023
% 0.14/0.36  % CPUTime    : 
% 3.58/3.85  SZS status Theorem for theBenchmark.p
% 3.58/3.85  SZS output start Proof for theBenchmark.p
% 3.58/3.85  Clause #0 (by assumption #[]): Eq
% 3.58/3.85    (Not
% 3.58/3.85      (∀ (Xf : a → a) (Xp : (a → a) → Prop),
% 3.58/3.85        And (Xp fun Xu => Xu) (∀ (Xj : a → a), Xp Xj → Xp fun Xx => Xf (Xj Xx)) → Xp fun Xx => Xf (Xf Xx)))
% 3.58/3.85    True
% 3.58/3.85  Clause #1 (by clausification #[0]): Eq
% 3.58/3.85    (∀ (Xf : a → a) (Xp : (a → a) → Prop),
% 3.58/3.85      And (Xp fun Xu => Xu) (∀ (Xj : a → a), Xp Xj → Xp fun Xx => Xf (Xj Xx)) → Xp fun Xx => Xf (Xf Xx))
% 3.58/3.85    False
% 3.58/3.85  Clause #2 (by clausification #[1]): ∀ (a_1 : a → a),
% 3.58/3.85    Eq
% 3.58/3.85      (Not
% 3.58/3.85        (∀ (Xp : (a → a) → Prop),
% 3.58/3.85          And (Xp fun Xu => Xu) (∀ (Xj : a → a), Xp Xj → Xp fun Xx => skS.0 0 a_1 (Xj Xx)) →
% 3.58/3.85            Xp fun Xx => skS.0 0 a_1 (skS.0 0 a_1 Xx)))
% 3.58/3.85      True
% 3.58/3.85  Clause #3 (by clausification #[2]): ∀ (a_1 : a → a),
% 3.58/3.85    Eq
% 3.58/3.85      (∀ (Xp : (a → a) → Prop),
% 3.58/3.85        And (Xp fun Xu => Xu) (∀ (Xj : a → a), Xp Xj → Xp fun Xx => skS.0 0 a_1 (Xj Xx)) →
% 3.58/3.85          Xp fun Xx => skS.0 0 a_1 (skS.0 0 a_1 Xx))
% 3.58/3.85      False
% 3.58/3.85  Clause #4 (by clausification #[3]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop),
% 3.58/3.85    Eq
% 3.58/3.85      (Not
% 3.58/3.85        (And (skS.0 1 a_1 a_2 fun Xu => Xu)
% 3.58/3.85            (∀ (Xj : a → a), skS.0 1 a_1 a_2 Xj → skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (Xj Xx)) →
% 3.58/3.85          skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (skS.0 0 a_1 Xx)))
% 3.58/3.85      True
% 3.58/3.85  Clause #5 (by clausification #[4]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop),
% 3.58/3.85    Eq
% 3.58/3.85      (And (skS.0 1 a_1 a_2 fun Xu => Xu)
% 3.58/3.85          (∀ (Xj : a → a), skS.0 1 a_1 a_2 Xj → skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (Xj Xx)) →
% 3.58/3.85        skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (skS.0 0 a_1 Xx))
% 3.58/3.85      False
% 3.58/3.85  Clause #6 (by clausification #[5]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop),
% 3.58/3.85    Eq
% 3.58/3.85      (And (skS.0 1 a_1 a_2 fun Xu => Xu)
% 3.58/3.85        (∀ (Xj : a → a), skS.0 1 a_1 a_2 Xj → skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (Xj Xx)))
% 3.58/3.85      True
% 3.58/3.85  Clause #7 (by clausification #[5]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop), Eq (skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (skS.0 0 a_1 Xx)) False
% 3.58/3.85  Clause #8 (by clausification #[6]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop),
% 3.58/3.85    Eq (∀ (Xj : a → a), skS.0 1 a_1 a_2 Xj → skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (Xj Xx)) True
% 3.58/3.85  Clause #9 (by clausification #[6]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop), Eq (skS.0 1 a_1 a_2 fun Xu => Xu) True
% 3.58/3.85  Clause #10 (by clausification #[8]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop) (a_3 : a → a),
% 3.58/3.85    Eq (skS.0 1 a_1 a_2 a_3 → skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (a_3 Xx)) True
% 3.58/3.85  Clause #11 (by clausification #[10]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop) (a_3 : a → a),
% 3.58/3.85    Or (Eq (skS.0 1 a_1 a_2 a_3) False) (Eq (skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (a_3 Xx)) True)
% 3.58/3.85  Clause #12 (by superposition #[9, 11]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop),
% 3.58/3.85    Or (Eq (skS.0 1 (fun x => a_1 x) (fun x => a_2 x) fun Xx => skS.0 0 (fun x => a_1 x) ((fun Xu => Xu) Xx)) True)
% 3.58/3.85      (Eq False True)
% 3.58/3.85  Clause #16 (by betaEtaReduce #[12]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop), Or (Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1)) True) (Eq False True)
% 3.58/3.85  Clause #17 (by clausification #[16]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1)) True
% 3.58/3.85  Clause #18 (by superposition #[17, 11]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop),
% 3.58/3.85    Or (Eq True False) (Eq (skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (skS.0 0 a_1 Xx)) True)
% 3.58/3.85  Clause #22 (by clausification #[18]): ∀ (a_1 : a → a) (a_2 : (a → a) → Prop), Eq (skS.0 1 a_1 a_2 fun Xx => skS.0 0 a_1 (skS.0 0 a_1 Xx)) True
% 3.58/3.85  Clause #23 (by superposition #[22, 7]): Eq True False
% 3.58/3.85  Clause #37 (by clausification #[23]): False
% 3.58/3.85  SZS output end Proof for theBenchmark.p
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