TSTP Solution File: ALG283^5 by Duper---1.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : ALG283^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n016.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 : Wed Aug 30 16:11:58 EDT 2023

% Result   : Theorem 3.98s 4.14s
% Output   : Proof 3.98s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem    : ALG283^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.13  % Command    : duper %s
% 0.18/0.34  % Computer : n016.cluster.edu
% 0.18/0.34  % Model    : x86_64 x86_64
% 0.18/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.18/0.34  % Memory   : 8042.1875MB
% 0.18/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.18/0.34  % CPULimit   : 300
% 0.18/0.34  % WCLimit    : 300
% 0.18/0.34  % DateTime   : Mon Aug 28 06:05:29 EDT 2023
% 0.18/0.35  % CPUTime    : 
% 3.98/4.14  SZS status Theorem for theBenchmark.p
% 3.98/4.14  SZS output start Proof for theBenchmark.p
% 3.98/4.14  Clause #0 (by assumption #[]): Eq
% 3.98/4.14    (Not
% 3.98/4.14      (And (And (∀ (Xx Xy Xz : a), Eq (cP (cP Xx Xy) Xz) (cP Xx (cP Xy Xz))) (∀ (Xx : a), Eq (cP cE Xx) Xx))
% 3.98/4.14          (∀ (Xy : a), Eq (cP (cJ Xy) Xy) cE) →
% 3.98/4.14        ∀ (X Y : a), Exists fun Z => Eq (cP X Z) Y))
% 3.98/4.14    True
% 3.98/4.14  Clause #1 (by clausification #[0]): Eq
% 3.98/4.14    (And (And (∀ (Xx Xy Xz : a), Eq (cP (cP Xx Xy) Xz) (cP Xx (cP Xy Xz))) (∀ (Xx : a), Eq (cP cE Xx) Xx))
% 3.98/4.14        (∀ (Xy : a), Eq (cP (cJ Xy) Xy) cE) →
% 3.98/4.14      ∀ (X Y : a), Exists fun Z => Eq (cP X Z) Y)
% 3.98/4.14    False
% 3.98/4.14  Clause #2 (by clausification #[1]): Eq
% 3.98/4.14    (And (And (∀ (Xx Xy Xz : a), Eq (cP (cP Xx Xy) Xz) (cP Xx (cP Xy Xz))) (∀ (Xx : a), Eq (cP cE Xx) Xx))
% 3.98/4.14      (∀ (Xy : a), Eq (cP (cJ Xy) Xy) cE))
% 3.98/4.14    True
% 3.98/4.14  Clause #3 (by clausification #[1]): Eq (∀ (X Y : a), Exists fun Z => Eq (cP X Z) Y) False
% 3.98/4.14  Clause #4 (by clausification #[2]): Eq (∀ (Xy : a), Eq (cP (cJ Xy) Xy) cE) True
% 3.98/4.14  Clause #5 (by clausification #[2]): Eq (And (∀ (Xx Xy Xz : a), Eq (cP (cP Xx Xy) Xz) (cP Xx (cP Xy Xz))) (∀ (Xx : a), Eq (cP cE Xx) Xx)) True
% 3.98/4.14  Clause #6 (by clausification #[4]): ∀ (a_1 : a), Eq (Eq (cP (cJ a_1) a_1) cE) True
% 3.98/4.14  Clause #7 (by clausification #[6]): ∀ (a_1 : a), Eq (cP (cJ a_1) a_1) cE
% 3.98/4.14  Clause #8 (by clausification #[3]): ∀ (a_1 : a), Eq (Not (∀ (Y : a), Exists fun Z => Eq (cP (skS.0 0 a_1) Z) Y)) True
% 3.98/4.14  Clause #9 (by clausification #[8]): ∀ (a_1 : a), Eq (∀ (Y : a), Exists fun Z => Eq (cP (skS.0 0 a_1) Z) Y) False
% 3.98/4.14  Clause #10 (by clausification #[9]): ∀ (a_1 a_2 : a), Eq (Not (Exists fun Z => Eq (cP (skS.0 0 a_1) Z) (skS.0 1 a_1 a_2))) True
% 3.98/4.14  Clause #11 (by clausification #[10]): ∀ (a_1 a_2 : a), Eq (Exists fun Z => Eq (cP (skS.0 0 a_1) Z) (skS.0 1 a_1 a_2)) False
% 3.98/4.14  Clause #12 (by clausification #[11]): ∀ (a_1 a_2 a_3 : a), Eq (Eq (cP (skS.0 0 a_1) a_2) (skS.0 1 a_1 a_3)) False
% 3.98/4.14  Clause #13 (by clausification #[12]): ∀ (a_1 a_2 a_3 : a), Ne (cP (skS.0 0 a_1) a_2) (skS.0 1 a_1 a_3)
% 3.98/4.14  Clause #14 (by clausification #[5]): Eq (∀ (Xx : a), Eq (cP cE Xx) Xx) True
% 3.98/4.14  Clause #15 (by clausification #[5]): Eq (∀ (Xx Xy Xz : a), Eq (cP (cP Xx Xy) Xz) (cP Xx (cP Xy Xz))) True
% 3.98/4.14  Clause #16 (by clausification #[14]): ∀ (a_1 : a), Eq (Eq (cP cE a_1) a_1) True
% 3.98/4.14  Clause #17 (by clausification #[16]): ∀ (a_1 : a), Eq (cP cE a_1) a_1
% 3.98/4.14  Clause #18 (by clausification #[15]): ∀ (a_1 : a), Eq (∀ (Xy Xz : a), Eq (cP (cP a_1 Xy) Xz) (cP a_1 (cP Xy Xz))) True
% 3.98/4.14  Clause #19 (by clausification #[18]): ∀ (a_1 a_2 : a), Eq (∀ (Xz : a), Eq (cP (cP a_1 a_2) Xz) (cP a_1 (cP a_2 Xz))) True
% 3.98/4.14  Clause #20 (by clausification #[19]): ∀ (a_1 a_2 a_3 : a), Eq (Eq (cP (cP a_1 a_2) a_3) (cP a_1 (cP a_2 a_3))) True
% 3.98/4.14  Clause #21 (by clausification #[20]): ∀ (a_1 a_2 a_3 : a), Eq (cP (cP a_1 a_2) a_3) (cP a_1 (cP a_2 a_3))
% 3.98/4.14  Clause #23 (by superposition #[21, 7]): ∀ (a_1 a_2 : a), Eq (cP cE a_1) (cP (cJ a_2) (cP a_2 a_1))
% 3.98/4.14  Clause #36 (by forward demodulation #[23, 17]): ∀ (a_1 a_2 : a), Eq a_1 (cP (cJ a_2) (cP a_2 a_1))
% 3.98/4.14  Clause #41 (by superposition #[36, 7]): ∀ (a_1 : a), Eq a_1 (cP (cJ (cJ a_1)) cE)
% 3.98/4.14  Clause #52 (by superposition #[41, 36]): ∀ (a_1 : a), Eq cE (cP (cJ (cJ (cJ a_1))) a_1)
% 3.98/4.14  Clause #110 (by superposition #[52, 36]): ∀ (a_1 : a), Eq a_1 (cP (cJ (cJ (cJ (cJ a_1)))) cE)
% 3.98/4.14  Clause #113 (by superposition #[110, 41]): ∀ (a_1 : a), Eq (cJ (cJ a_1)) a_1
% 3.98/4.14  Clause #122 (by superposition #[113, 36]): ∀ (a_1 a_2 : a), Eq a_1 (cP a_2 (cP (cJ a_2) a_1))
% 3.98/4.14  Clause #180 (by superposition #[122, 13]): ∀ (a_1 a_2 a_3 : a), Ne a_1 (skS.0 1 a_2 a_3)
% 3.98/4.14  Clause #234 (by destructive equality resolution #[180]): False
% 3.98/4.14  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------