TSTP Solution File: SEU920^5 by Duper---1.0

View Problem - Process Solution

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% File     : Duper---1.0
% Problem  : SEU920^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n005.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 16:44:18 EDT 2023

% Result   : Theorem 3.31s 3.61s
% Output   : Proof 3.31s
% 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    : SEU920^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14  % Command    : duper %s
% 0.16/0.35  % Computer : n005.cluster.edu
% 0.16/0.35  % Model    : x86_64 x86_64
% 0.16/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.16/0.35  % Memory   : 8042.1875MB
% 0.16/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.16/0.35  % CPULimit   : 300
% 0.16/0.35  % WCLimit    : 300
% 0.16/0.35  % DateTime   : Wed Aug 23 18:38:09 EDT 2023
% 0.16/0.35  % CPUTime    : 
% 3.31/3.61  SZS status Theorem for theBenchmark.p
% 3.31/3.61  SZS output start Proof for theBenchmark.p
% 3.31/3.61  Clause #0 (by assumption #[]): Eq
% 3.31/3.61    (Not (∀ (F : a → b) (G : b → c), (∀ (Y : c), Exists fun X => Eq (G (F X)) Y) → ∀ (Y : c), Exists fun X => Eq (G X) Y))
% 3.31/3.61    True
% 3.31/3.61  Clause #1 (by clausification #[0]): Eq (∀ (F : a → b) (G : b → c), (∀ (Y : c), Exists fun X => Eq (G (F X)) Y) → ∀ (Y : c), Exists fun X => Eq (G X) Y)
% 3.31/3.61    False
% 3.31/3.61  Clause #2 (by clausification #[1]): ∀ (a_1 : a → b),
% 3.31/3.61    Eq
% 3.31/3.61      (Not (∀ (G : b → c), (∀ (Y : c), Exists fun X => Eq (G (skS.0 0 a_1 X)) Y) → ∀ (Y : c), Exists fun X => Eq (G X) Y))
% 3.31/3.61      True
% 3.31/3.61  Clause #3 (by clausification #[2]): ∀ (a_1 : a → b),
% 3.31/3.61    Eq (∀ (G : b → c), (∀ (Y : c), Exists fun X => Eq (G (skS.0 0 a_1 X)) Y) → ∀ (Y : c), Exists fun X => Eq (G X) Y)
% 3.31/3.61      False
% 3.31/3.61  Clause #4 (by clausification #[3]): ∀ (a_1 : a → b) (a_2 : b → c),
% 3.31/3.61    Eq
% 3.31/3.61      (Not
% 3.31/3.61        ((∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 X)) Y) →
% 3.31/3.61          ∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 X) Y))
% 3.31/3.61      True
% 3.31/3.61  Clause #5 (by clausification #[4]): ∀ (a_1 : a → b) (a_2 : b → c),
% 3.31/3.61    Eq
% 3.31/3.61      ((∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 X)) Y) →
% 3.31/3.61        ∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 X) Y)
% 3.31/3.61      False
% 3.31/3.61  Clause #6 (by clausification #[5]): ∀ (a_1 : a → b) (a_2 : b → c), Eq (∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 X)) Y) True
% 3.31/3.61  Clause #7 (by clausification #[5]): ∀ (a_1 : a → b) (a_2 : b → c), Eq (∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 X) Y) False
% 3.31/3.61  Clause #8 (by clausification #[6]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : c), Eq (Exists fun X => Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 X)) a_3) True
% 3.31/3.61  Clause #9 (by clausification #[8]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : c) (a_4 : a),
% 3.31/3.61    Eq (Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3 a_4))) a_3) True
% 3.31/3.61  Clause #10 (by clausification #[9]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : c) (a_4 : a), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3 a_4))) a_3
% 3.31/3.61  Clause #11 (by clausification #[7]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : c), Eq (Not (Exists fun X => Eq (skS.0 1 a_1 a_2 X) (skS.0 3 a_1 a_2 a_3))) True
% 3.31/3.61  Clause #12 (by clausification #[11]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : c), Eq (Exists fun X => Eq (skS.0 1 a_1 a_2 X) (skS.0 3 a_1 a_2 a_3)) False
% 3.31/3.61  Clause #13 (by clausification #[12]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : b) (a_4 : c), Eq (Eq (skS.0 1 a_1 a_2 a_3) (skS.0 3 a_1 a_2 a_4)) False
% 3.31/3.61  Clause #14 (by clausification #[13]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : b) (a_4 : c), Ne (skS.0 1 a_1 a_2 a_3) (skS.0 3 a_1 a_2 a_4)
% 3.31/3.61  Clause #15 (by superposition #[14, 10]): ∀ (a_1 : c) (a_2 : a → b) (a_3 : b → c) (a_4 : c), Ne a_1 (skS.0 3 a_2 a_3 a_4)
% 3.31/3.61  Clause #16 (by destructive equality resolution #[15]): False
% 3.31/3.61  SZS output end Proof for theBenchmark.p
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