↑ Up

Duper---1.0.THM-Prf.s

View TPTP
Problem
Process solution in
SystemOnTSTP
Download .tgz
%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SEU568^2 : TPTP v9.2.0. Released v3.7.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n004.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 08:01:56 PM UTC 2025

% Result   : Theorem 3.58s 3.75s
% Output   : Proof 3.58s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12  % Problem    : SEU568^2 : TPTP v9.2.0. Released v3.7.0.
% 0.11/0.13  % Command    : duper %s
% 0.13/0.34  % Computer : n004.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit   : 300
% 0.13/0.34  % WCLimit    : 300
% 0.13/0.34  % DateTime   : Fri Oct  3 06:12:53 EDT 2025
% 0.13/0.34  % CPUTime    : 
% 3.58/3.75  SZS status Theorem for theBenchmark.p
% 3.58/3.75  SZS output start Proof for theBenchmark.p
% 3.58/3.75  Clause #2 (by assumption #[]): Eq (Not (notsubsetI → notequalI1 → ∀ (A B Xx : Iota), in Xx A → Not (in Xx B) → Ne A B)) True
% 3.58/3.75  Clause #24 (by clausification #[2]): Eq (notsubsetI → notequalI1 → ∀ (A B Xx : Iota), in Xx A → Not (in Xx B) → Ne A B) False
% 3.58/3.75  Clause #26 (by clausification #[24]): Eq (notequalI1 → ∀ (A B Xx : Iota), in Xx A → Not (in Xx B) → Ne A B) False
% 3.58/3.75  Clause #36 (by clausification #[26]): Eq (∀ (A B Xx : Iota), in Xx A → Not (in Xx B) → Ne A B) False
% 3.58/3.75  Clause #46 (by clausification #[36]): ∀ (a : Iota), Eq (Not (∀ (B Xx : Iota), in Xx (skS.0 2 a) → Not (in Xx B) → Ne (skS.0 2 a) B)) True
% 3.58/3.75  Clause #47 (by clausification #[46]): ∀ (a : Iota), Eq (∀ (B Xx : Iota), in Xx (skS.0 2 a) → Not (in Xx B) → Ne (skS.0 2 a) B) False
% 3.58/3.75  Clause #48 (by clausification #[47]): ∀ (a a_1 : Iota),
% 3.58/3.75    Eq (Not (∀ (Xx : Iota), in Xx (skS.0 2 a) → Not (in Xx (skS.0 3 a a_1)) → Ne (skS.0 2 a) (skS.0 3 a a_1))) True
% 3.58/3.75  Clause #49 (by clausification #[48]): ∀ (a a_1 : Iota),
% 3.58/3.75    Eq (∀ (Xx : Iota), in Xx (skS.0 2 a) → Not (in Xx (skS.0 3 a a_1)) → Ne (skS.0 2 a) (skS.0 3 a a_1)) False
% 3.58/3.75  Clause #50 (by clausification #[49]): ∀ (a a_1 a_2 : Iota),
% 3.58/3.75    Eq
% 3.58/3.75      (Not
% 3.58/3.75        (in (skS.0 4 a a_1 a_2) (skS.0 2 a) →
% 3.58/3.75          Not (in (skS.0 4 a a_1 a_2) (skS.0 3 a a_1)) → Ne (skS.0 2 a) (skS.0 3 a a_1)))
% 3.58/3.75      True
% 3.58/3.75  Clause #51 (by clausification #[50]): ∀ (a a_1 a_2 : Iota),
% 3.58/3.75    Eq
% 3.58/3.75      (in (skS.0 4 a a_1 a_2) (skS.0 2 a) → Not (in (skS.0 4 a a_1 a_2) (skS.0 3 a a_1)) → Ne (skS.0 2 a) (skS.0 3 a a_1))
% 3.58/3.75      False
% 3.58/3.75  Clause #52 (by clausification #[51]): ∀ (a a_1 a_2 : Iota), Eq (in (skS.0 4 a a_1 a_2) (skS.0 2 a)) True
% 3.58/3.75  Clause #53 (by clausification #[51]): ∀ (a a_1 a_2 : Iota), Eq (Not (in (skS.0 4 a a_1 a_2) (skS.0 3 a a_1)) → Ne (skS.0 2 a) (skS.0 3 a a_1)) False
% 3.58/3.75  Clause #74 (by clausification #[53]): ∀ (a a_1 a_2 : Iota), Eq (Not (in (skS.0 4 a a_1 a_2) (skS.0 3 a a_1))) True
% 3.58/3.75  Clause #75 (by clausification #[53]): ∀ (a a_1 : Iota), Eq (Ne (skS.0 2 a) (skS.0 3 a a_1)) False
% 3.58/3.75  Clause #76 (by clausification #[74]): ∀ (a a_1 a_2 : Iota), Eq (in (skS.0 4 a a_1 a_2) (skS.0 3 a a_1)) False
% 3.58/3.75  Clause #77 (by clausification #[75]): ∀ (a a_1 : Iota), Eq (skS.0 2 a) (skS.0 3 a a_1)
% 3.58/3.75  Clause #78 (by backward demodulation #[77, 76]): ∀ (a a_1 a_2 : Iota), Eq (in (skS.0 4 a a_1 a_2) (skS.0 2 a)) False
% 3.58/3.75  Clause #79 (by superposition #[78, 52]): Eq False True
% 3.58/3.75  Clause #80 (by clausification #[79]): False
% 3.58/3.75  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------