TSTP Solution File: SEU834^5 by Duper---1.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SEU834^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n022.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 16:44:06 EDT 2023

% Result   : Theorem 3.54s 3.74s
% Output   : Proof 3.54s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem    : SEU834^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14  % Command    : duper %s
% 0.13/0.35  % Computer : n022.cluster.edu
% 0.13/0.35  % Model    : x86_64 x86_64
% 0.13/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35  % Memory   : 8042.1875MB
% 0.13/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35  % CPULimit   : 300
% 0.13/0.35  % WCLimit    : 300
% 0.13/0.35  % DateTime   : Wed Aug 23 12:17:26 EDT 2023
% 0.13/0.35  % CPUTime    : 
% 3.54/3.74  SZS status Theorem for theBenchmark.p
% 3.54/3.74  SZS output start Proof for theBenchmark.p
% 3.54/3.74  Clause #0 (by assumption #[]): Eq (Not (∀ (S T : a → Prop) (Xx : a), And (S Xx) (T Xx) → T Xx)) True
% 3.54/3.74  Clause #1 (by clausification #[0]): Eq (∀ (S T : a → Prop) (Xx : a), And (S Xx) (T Xx) → T Xx) False
% 3.54/3.74  Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop), Eq (Not (∀ (T : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (T Xx) → T Xx)) True
% 3.54/3.74  Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop), Eq (∀ (T : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (T Xx) → T Xx) False
% 3.54/3.74  Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop), Eq (Not (∀ (Xx : a), And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → skS.0 1 a_1 a_2 Xx)) True
% 3.54/3.74  Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop), Eq (∀ (Xx : a), And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → skS.0 1 a_1 a_2 Xx) False
% 3.54/3.74  Clause #6 (by clausification #[5]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.54/3.74    Eq
% 3.54/3.74      (Not
% 3.54/3.74        (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)) →
% 3.54/3.74          skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)))
% 3.54/3.74      True
% 3.54/3.74  Clause #7 (by clausification #[6]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.54/3.74    Eq
% 3.54/3.74      (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)) →
% 3.54/3.74        skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))
% 3.54/3.74      False
% 3.54/3.74  Clause #8 (by clausification #[7]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.54/3.74    Eq (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) True
% 3.54/3.74  Clause #9 (by clausification #[7]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)) False
% 3.54/3.74  Clause #10 (by clausification #[8]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)) True
% 3.54/3.74  Clause #12 (by superposition #[9, 10]): Eq True False
% 3.54/3.74  Clause #13 (by clausification #[12]): False
% 3.54/3.74  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------