↑ Up

Duper---1.0.THM-Prf.s

View TPTP
Problem
Process solution in
SystemOnTSTP
Download .tgz
%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : NUM647^1 : TPTP v9.2.0. Released v3.7.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n028.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 07:56:55 PM UTC 2025

% Result   : Theorem 5.19s 5.42s
% Output   : Proof 5.19s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.11  % Problem    : NUM647^1 : TPTP v9.2.0. Released v3.7.0.
% 0.07/0.12  % Command    : duper %s
% 0.11/0.31  % Computer : n028.cluster.edu
% 0.11/0.31  % Model    : x86_64 x86_64
% 0.11/0.31  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.31  % Memory   : 8042.1875MB
% 0.11/0.31  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.11/0.31  % CPULimit   : 300
% 0.11/0.31  % WCLimit    : 300
% 0.11/0.31  % DateTime   : Fri Oct  3 06:51:23 EDT 2025
% 0.15/0.31  % CPUTime    : 
% 5.19/5.42  SZS status Theorem for theBenchmark.p
% 5.19/5.42  SZS output start Proof for theBenchmark.p
% 5.19/5.42  Clause #0 (by assumption #[]): Eq (Eq (pl x y) (pl x z)) True
% 5.19/5.42  Clause #2 (by assumption #[]): Eq (∀ (Xx Xy Xz : nat), Ne Xy Xz → Ne (pl Xx Xy) (pl Xx Xz)) True
% 5.19/5.42  Clause #3 (by assumption #[]): Eq (Not (Eq y z)) True
% 5.19/5.42  Clause #8 (by clausification #[3]): Eq (Eq y z) False
% 5.19/5.42  Clause #9 (by clausification #[8]): Ne y z
% 5.19/5.42  Clause #10 (by clausification #[0]): Eq (pl x y) (pl x z)
% 5.19/5.42  Clause #11 (by clausification #[2]): ∀ (a : nat), Eq (∀ (Xy Xz : nat), Ne Xy Xz → Ne (pl a Xy) (pl a Xz)) True
% 5.19/5.42  Clause #12 (by clausification #[11]): ∀ (a a_1 : nat), Eq (∀ (Xz : nat), Ne a Xz → Ne (pl a_1 a) (pl a_1 Xz)) True
% 5.19/5.42  Clause #13 (by clausification #[12]): ∀ (a a_1 a_2 : nat), Eq (Ne a a_1 → Ne (pl a_2 a) (pl a_2 a_1)) True
% 5.19/5.42  Clause #14 (by clausification #[13]): ∀ (a a_1 a_2 : nat), Or (Eq (Ne a a_1) False) (Eq (Ne (pl a_2 a) (pl a_2 a_1)) True)
% 5.19/5.42  Clause #15 (by clausification #[14]): ∀ (a a_1 a_2 : nat), Or (Eq (Ne (pl a a_1) (pl a a_2)) True) (Eq a_1 a_2)
% 5.19/5.42  Clause #16 (by clausification #[15]): ∀ (a a_1 a_2 : nat), Or (Eq a a_1) (Ne (pl a_2 a) (pl a_2 a_1))
% 5.19/5.42  Clause #18 (by superposition #[16, 10]): ∀ (a : nat), Or (Eq y a) (Ne (pl x z) (pl x a))
% 5.19/5.42  Clause #20 (by equality resolution #[18]): Eq y z
% 5.19/5.42  Clause #21 (by forward contextual literal cutting #[20, 9]): False
% 5.19/5.42  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------