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Duper---1.0.THM-Prf.s

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%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SEV134^5 : TPTP v9.2.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n012.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 : Fri Oct  3 08:02:52 PM UTC 2025

% Result   : Theorem 4.07s 4.27s
% Output   : Proof 4.07s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.16  % Problem    : SEV134^5 : TPTP v9.2.0. Released v4.0.0.
% 0.12/0.17  % Command    : duper %s
% 0.16/0.39  % Computer : n012.cluster.edu
% 0.16/0.39  % Model    : x86_64 x86_64
% 0.16/0.39  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.16/0.39  % Memory   : 8042.1875MB
% 0.16/0.39  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.16/0.39  % CPULimit   : 300
% 0.16/0.39  % WCLimit    : 300
% 0.16/0.39  % DateTime   : Fri Oct  3 01:08:53 EDT 2025
% 0.16/0.39  % CPUTime    : 
% 4.07/4.27  SZS status Theorem for theBenchmark.p
% 4.07/4.27  SZS output start Proof for theBenchmark.p
% 4.07/4.27  Clause #0 (by assumption #[]): Eq
% 4.07/4.27    (Not
% 4.07/4.27      (∀ (Xr : a → a → Prop) (Xx : a) (Xx0 : a → Prop),
% 4.07/4.27        (∀ (Xy Xz : a), And (Xr Xy Xz) (Xx0 Xy) → Xx0 Xz) → Xx0 Xx → Xx0 Xx))
% 4.07/4.27    True
% 4.07/4.27  Clause #1 (by clausification #[0]): Eq
% 4.07/4.27    (∀ (Xr : a → a → Prop) (Xx : a) (Xx0 : a → Prop), (∀ (Xy Xz : a), And (Xr Xy Xz) (Xx0 Xy) → Xx0 Xz) → Xx0 Xx → Xx0 Xx)
% 4.07/4.27    False
% 4.07/4.27  Clause #2 (by clausification #[1]): ∀ (a_1 : a → a → Prop),
% 4.07/4.27    Eq (Not (∀ (Xx : a) (Xx0 : a → Prop), (∀ (Xy Xz : a), And (skS.0 0 a_1 Xy Xz) (Xx0 Xy) → Xx0 Xz) → Xx0 Xx → Xx0 Xx))
% 4.07/4.27      True
% 4.07/4.27  Clause #3 (by clausification #[2]): ∀ (a_1 : a → a → Prop),
% 4.07/4.27    Eq (∀ (Xx : a) (Xx0 : a → Prop), (∀ (Xy Xz : a), And (skS.0 0 a_1 Xy Xz) (Xx0 Xy) → Xx0 Xz) → Xx0 Xx → Xx0 Xx) False
% 4.07/4.27  Clause #4 (by clausification #[3]): ∀ (a_1 : a → a → Prop) (a_2 : a),
% 4.07/4.27    Eq
% 4.07/4.27      (Not
% 4.07/4.27        (∀ (Xx0 : a → Prop),
% 4.07/4.27          (∀ (Xy Xz : a), And (skS.0 0 a_1 Xy Xz) (Xx0 Xy) → Xx0 Xz) → Xx0 (skS.0 1 a_1 a_2) → Xx0 (skS.0 1 a_1 a_2)))
% 4.07/4.27      True
% 4.07/4.27  Clause #5 (by clausification #[4]): ∀ (a_1 : a → a → Prop) (a_2 : a),
% 4.07/4.27    Eq
% 4.07/4.27      (∀ (Xx0 : a → Prop),
% 4.07/4.27        (∀ (Xy Xz : a), And (skS.0 0 a_1 Xy Xz) (Xx0 Xy) → Xx0 Xz) → Xx0 (skS.0 1 a_1 a_2) → Xx0 (skS.0 1 a_1 a_2))
% 4.07/4.27      False
% 4.07/4.27  Clause #6 (by clausification #[5]): ∀ (a_1 : a → a → Prop) (a_2 : a) (a_3 : a → Prop),
% 4.07/4.27    Eq
% 4.07/4.27      (Not
% 4.07/4.27        ((∀ (Xy Xz : a), And (skS.0 0 a_1 Xy Xz) (skS.0 2 a_1 a_2 a_3 Xy) → skS.0 2 a_1 a_2 a_3 Xz) →
% 4.07/4.27          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 (skS.0 1 a_1 a_2)))
% 4.07/4.27      True
% 4.07/4.27  Clause #7 (by clausification #[6]): ∀ (a_1 : a → a → Prop) (a_2 : a) (a_3 : a → Prop),
% 4.07/4.27    Eq
% 4.07/4.27      ((∀ (Xy Xz : a), And (skS.0 0 a_1 Xy Xz) (skS.0 2 a_1 a_2 a_3 Xy) → skS.0 2 a_1 a_2 a_3 Xz) →
% 4.07/4.27        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 (skS.0 1 a_1 a_2))
% 4.07/4.27      False
% 4.07/4.27  Clause #9 (by clausification #[7]): ∀ (a_1 : a → a → Prop) (a_2 : a) (a_3 : a → Prop),
% 4.07/4.27    Eq (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 (skS.0 1 a_1 a_2)) False
% 4.07/4.27  Clause #14 (by clausification #[9]): ∀ (a_1 : a → a → Prop) (a_2 : a) (a_3 : a → Prop), Eq (skS.0 2 a_1 a_2 a_3 (skS.0 1 a_1 a_2)) True
% 4.07/4.27  Clause #15 (by clausification #[9]): ∀ (a_1 : a → a → Prop) (a_2 : a) (a_3 : a → Prop), Eq (skS.0 2 a_1 a_2 a_3 (skS.0 1 a_1 a_2)) False
% 4.07/4.27  Clause #16 (by superposition #[15, 14]): Eq False True
% 4.07/4.27  Clause #17 (by clausification #[16]): False
% 4.07/4.27  SZS output end Proof for theBenchmark.p
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