↑ Up

Duper---1.0.THM-Prf.s

View TPTP
Problem
Process solution in
SystemOnTSTP
Download .tgz
%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : NUM709^1 : TPTP v9.2.0. Released v3.7.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n008.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 07:57:09 PM UTC 2025

% Result   : Theorem 3.42s 3.66s
% Output   : Proof 3.42s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.12  % Problem    : NUM709^1 : TPTP v9.2.0. Released v3.7.0.
% 0.08/0.13  % Command    : duper %s
% 0.13/0.35  % Computer : n008.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   : Fri Oct  3 07:59:23 EDT 2025
% 0.13/0.35  % CPUTime    : 
% 3.42/3.66  SZS status Theorem for theBenchmark.p
% 3.42/3.66  SZS output start Proof for theBenchmark.p
% 3.42/3.66  Clause #0 (by assumption #[]): Eq (∀ (Xx Xy : nat), Eq (ts (suc Xx) Xy) (pl (ts Xx Xy) Xy)) True
% 3.42/3.66  Clause #1 (by assumption #[]): Eq (Not (Eq (pl (ts x y) y) (ts (suc x) y))) True
% 3.42/3.66  Clause #2 (by clausification #[1]): Eq (Eq (pl (ts x y) y) (ts (suc x) y)) False
% 3.42/3.66  Clause #3 (by clausification #[2]): Ne (pl (ts x y) y) (ts (suc x) y)
% 3.42/3.66  Clause #4 (by clausification #[0]): ∀ (a : nat), Eq (∀ (Xy : nat), Eq (ts (suc a) Xy) (pl (ts a Xy) Xy)) True
% 3.42/3.66  Clause #5 (by clausification #[4]): ∀ (a a_1 : nat), Eq (Eq (ts (suc a) a_1) (pl (ts a a_1) a_1)) True
% 3.42/3.66  Clause #6 (by clausification #[5]): ∀ (a a_1 : nat), Eq (ts (suc a) a_1) (pl (ts a a_1) a_1)
% 3.42/3.66  Clause #8 (by backward contextual literal cutting #[6, 3]): False
% 3.42/3.66  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------