↑ Up

Duper---1.0.THM-Prf.s

View TPTP
Problem
Process solution in
SystemOnTSTP
Download .tgz
%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : NUM771^1 : TPTP v9.2.0. Released v3.7.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n015.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 07:57:18 PM UTC 2025

% Result   : Theorem 3.32s 3.49s
% Output   : Proof 3.32s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12  % Problem    : NUM771^1 : TPTP v9.2.0. Released v3.7.0.
% 0.06/0.12  % Command    : duper %s
% 0.12/0.34  % Computer : n015.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34  % CPULimit   : 300
% 0.12/0.34  % WCLimit    : 300
% 0.12/0.34  % DateTime   : Fri Oct  3 07:42:38 EDT 2025
% 0.12/0.34  % CPUTime    : 
% 3.32/3.49  SZS status Theorem for theBenchmark.p
% 3.32/3.49  SZS output start Proof for theBenchmark.p
% 3.32/3.49  Clause #0 (by assumption #[]): Eq (∀ (Xx : frac), eq Xx Xx) True
% 3.32/3.49  Clause #1 (by assumption #[]): Eq (∀ (Xx Xy : nat), Eq (ts Xx Xy) (ts Xy Xx)) True
% 3.32/3.49  Clause #2 (by assumption #[]): Eq (Not (eq (fr (ts (num x) (num y)) (ts (den x) (den y))) (fr (ts (num y) (num x)) (ts (den y) (den x))))) True
% 3.32/3.49  Clause #3 (by clausification #[0]): ∀ (a : frac), Eq (eq a a) True
% 3.32/3.49  Clause #4 (by clausification #[1]): ∀ (a : nat), Eq (∀ (Xy : nat), Eq (ts a Xy) (ts Xy a)) True
% 3.32/3.49  Clause #5 (by clausification #[4]): ∀ (a a_1 : nat), Eq (Eq (ts a a_1) (ts a_1 a)) True
% 3.32/3.49  Clause #6 (by clausification #[5]): ∀ (a a_1 : nat), Eq (ts a a_1) (ts a_1 a)
% 3.32/3.49  Clause #7 (by clausification #[2]): Eq (eq (fr (ts (num x) (num y)) (ts (den x) (den y))) (fr (ts (num y) (num x)) (ts (den y) (den x)))) False
% 3.32/3.49  Clause #8 (by forward demodulation #[7, 6]): Eq (eq (fr (ts (num x) (num y)) (ts (den y) (den x))) (fr (ts (num y) (num x)) (ts (den y) (den x)))) False
% 3.32/3.49  Clause #9 (by forward demodulation #[8, 6]): Eq (eq (fr (ts (num y) (num x)) (ts (den y) (den x))) (fr (ts (num y) (num x)) (ts (den y) (den x)))) False
% 3.32/3.49  Clause #10 (by superposition #[9, 3]): Eq False True
% 3.32/3.49  Clause #11 (by clausification #[10]): False
% 3.32/3.49  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------