TSTP Solution File: SET631^5 by Duper---1.0

View Problem - Process Solution

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% File     : Duper---1.0
% Problem  : SET631^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n027.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 14:47:11 EDT 2023

% Result   : Theorem 3.73s 4.00s
% Output   : Proof 3.73s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.14  % Problem    : SET631^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.15  % Command    : duper %s
% 0.14/0.37  % Computer : n027.cluster.edu
% 0.14/0.37  % Model    : x86_64 x86_64
% 0.14/0.37  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.37  % Memory   : 8042.1875MB
% 0.14/0.37  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.37  % CPULimit   : 300
% 0.14/0.37  % WCLimit    : 300
% 0.14/0.37  % DateTime   : Sat Aug 26 08:47:34 EDT 2023
% 0.22/0.37  % CPUTime    : 
% 3.73/4.00  SZS status Theorem for theBenchmark.p
% 3.73/4.00  SZS output start Proof for theBenchmark.p
% 3.73/4.00  Clause #0 (by assumption #[]): Eq
% 3.73/4.00    (Not
% 3.73/4.00      (∀ (X Y Z : a → Prop),
% 3.73/4.00        (Exists fun Xx => And (And (X Xx) (Y Xx)) (Not (Z Xx))) → Exists fun Xx => And (X Xx) (Y Xx)))
% 3.73/4.00    True
% 3.73/4.00  Clause #1 (by clausification #[0]): Eq (∀ (X Y Z : a → Prop), (Exists fun Xx => And (And (X Xx) (Y Xx)) (Not (Z Xx))) → Exists fun Xx => And (X Xx) (Y Xx))
% 3.73/4.00    False
% 3.73/4.00  Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop),
% 3.73/4.00    Eq
% 3.73/4.00      (Not
% 3.73/4.00        (∀ (Y Z : a → Prop),
% 3.73/4.00          (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (Not (Z Xx))) →
% 3.73/4.00            Exists fun Xx => And (skS.0 0 a_1 Xx) (Y Xx)))
% 3.73/4.00      True
% 3.73/4.00  Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop),
% 3.73/4.00    Eq
% 3.73/4.00      (∀ (Y Z : a → Prop),
% 3.73/4.00        (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (Not (Z Xx))) → Exists fun Xx => And (skS.0 0 a_1 Xx) (Y Xx))
% 3.73/4.00      False
% 3.73/4.00  Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 3.73/4.00    Eq
% 3.73/4.00      (Not
% 3.73/4.00        (∀ (Z : a → Prop),
% 3.73/4.00          (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (Z Xx))) →
% 3.73/4.00            Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)))
% 3.73/4.00      True
% 3.73/4.00  Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 3.73/4.00    Eq
% 3.73/4.00      (∀ (Z : a → Prop),
% 3.73/4.00        (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (Z Xx))) →
% 3.73/4.00          Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx))
% 3.73/4.00      False
% 3.73/4.00  Clause #6 (by clausification #[5]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.73/4.00    Eq
% 3.73/4.00      (Not
% 3.73/4.00        ((Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))) →
% 3.73/4.00          Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)))
% 3.73/4.00      True
% 3.73/4.00  Clause #7 (by clausification #[6]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.73/4.00    Eq
% 3.73/4.00      ((Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))) →
% 3.73/4.00        Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx))
% 3.73/4.00      False
% 3.73/4.00  Clause #8 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.73/4.00    Eq (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))) True
% 3.73/4.00  Clause #9 (by clausification #[7]): ∀ (a_1 a_2 : a → Prop), Eq (Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) False
% 3.73/4.00  Clause #10 (by clausification #[8]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a),
% 3.73/4.00    Eq
% 3.73/4.00      (And (And (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3 a_4)))
% 3.73/4.00        (Not (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4))))
% 3.73/4.00      True
% 3.73/4.00  Clause #12 (by clausification #[10]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a),
% 3.73/4.00    Eq (And (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3 a_4))) True
% 3.73/4.00  Clause #14 (by clausification #[9]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 : a → Prop), Eq (And (skS.0 0 a_1 a_2) (skS.0 1 a_1 a_3 a_2)) False
% 3.73/4.00  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) False)
% 3.73/4.00  Clause #16 (by clausification #[12]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3 a_4)) True
% 3.73/4.00  Clause #17 (by clausification #[12]): ∀ (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
% 3.73/4.00  Clause #18 (by superposition #[17, 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)) False)
% 3.73/4.00  Clause #19 (by clausification #[18]): ∀ (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)) False
% 3.73/4.00  Clause #20 (by superposition #[19, 16]): Eq False True
% 3.73/4.00  Clause #21 (by clausification #[20]): False
% 3.73/4.00  SZS output end Proof for theBenchmark.p
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