↑ Up

Duper---1.0.THM-Prf.s

View TPTP
Problem
Process solution in
SystemOnTSTP
Download .tgz
%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SYN325+1 : TPTP v9.2.0. Released v2.0.0.
% 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 08:07:24 PM UTC 2025

% Result   : Theorem 5.72s 5.97s
% Output   : Proof 5.72s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.09/0.11  % Problem    : SYN325+1 : TPTP v9.2.0. Released v2.0.0.
% 0.09/0.12  % Command    : duper %s
% 0.12/0.33  % Computer : n015.cluster.edu
% 0.12/0.33  % Model    : x86_64 x86_64
% 0.12/0.33  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33  % Memory   : 8042.1875MB
% 0.12/0.33  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33  % CPULimit   : 300
% 0.12/0.33  % WCLimit    : 300
% 0.12/0.33  % DateTime   : Thu Oct  2 15:31:38 EDT 2025
% 0.12/0.33  % CPUTime    : 
% 5.72/5.97  SZS status Theorem for theBenchmark.p
% 5.72/5.97  SZS output start Proof for theBenchmark.p
% 5.72/5.97  Clause #0 (by assumption #[]): Eq (Not (Exists fun X => ∀ (Y : Iota), ((big_f X X → big_f Y Y) → And (big_f X Y) (big_g X)) → big_g Y)) True
% 5.72/5.97  Clause #1 (by clausification #[0]): Eq (Exists fun X => ∀ (Y : Iota), ((big_f X X → big_f Y Y) → And (big_f X Y) (big_g X)) → big_g Y) False
% 5.72/5.97  Clause #2 (by clausification #[1]): ∀ (a : Iota), Eq (∀ (Y : Iota), ((big_f a a → big_f Y Y) → And (big_f a Y) (big_g a)) → big_g Y) False
% 5.72/5.97  Clause #3 (by clausification #[2]): ∀ (a a_1 : Iota),
% 5.72/5.97    Eq
% 5.72/5.97      (Not
% 5.72/5.97        (((big_f a a → big_f (skS.0 0 a a_1) (skS.0 0 a a_1)) → And (big_f a (skS.0 0 a a_1)) (big_g a)) →
% 5.72/5.97          big_g (skS.0 0 a a_1)))
% 5.72/5.97      True
% 5.72/5.97  Clause #4 (by clausification #[3]): ∀ (a a_1 : Iota),
% 5.72/5.97    Eq
% 5.72/5.97      (((big_f a a → big_f (skS.0 0 a a_1) (skS.0 0 a a_1)) → And (big_f a (skS.0 0 a a_1)) (big_g a)) →
% 5.72/5.97        big_g (skS.0 0 a a_1))
% 5.72/5.97      False
% 5.72/5.97  Clause #5 (by clausification #[4]): ∀ (a a_1 : Iota),
% 5.72/5.97    Eq ((big_f a a → big_f (skS.0 0 a a_1) (skS.0 0 a a_1)) → And (big_f a (skS.0 0 a a_1)) (big_g a)) True
% 5.72/5.97  Clause #6 (by clausification #[4]): ∀ (a a_1 : Iota), Eq (big_g (skS.0 0 a a_1)) False
% 5.72/5.97  Clause #7 (by clausification #[5]): ∀ (a a_1 : Iota),
% 5.72/5.97    Or (Eq (big_f a a → big_f (skS.0 0 a a_1) (skS.0 0 a a_1)) False) (Eq (And (big_f a (skS.0 0 a a_1)) (big_g a)) True)
% 5.72/5.97  Clause #8 (by clausification #[7]): ∀ (a a_1 : Iota), Or (Eq (And (big_f a (skS.0 0 a a_1)) (big_g a)) True) (Eq (big_f a a) True)
% 5.72/5.97  Clause #9 (by clausification #[7]): ∀ (a a_1 : Iota),
% 5.72/5.97    Or (Eq (And (big_f a (skS.0 0 a a_1)) (big_g a)) True) (Eq (big_f (skS.0 0 a a_1) (skS.0 0 a a_1)) False)
% 5.72/5.97  Clause #10 (by clausification #[8]): ∀ (a : Iota), Or (Eq (big_f a a) True) (Eq (big_g a) True)
% 5.72/5.97  Clause #12 (by clausification #[9]): ∀ (a a_1 : Iota), Or (Eq (big_f (skS.0 0 a a_1) (skS.0 0 a a_1)) False) (Eq (big_g a) True)
% 5.72/5.97  Clause #14 (by superposition #[12, 10]): ∀ (a a_1 : Iota), Or (Eq (big_g a) True) (Or (Eq False True) (Eq (big_g (skS.0 0 a a_1)) True))
% 5.72/5.97  Clause #15 (by clausification #[14]): ∀ (a a_1 : Iota), Or (Eq (big_g a) True) (Eq (big_g (skS.0 0 a a_1)) True)
% 5.72/5.97  Clause #16 (by superposition #[15, 6]): ∀ (a : Iota), Or (Eq (big_g a) True) (Eq True False)
% 5.72/5.97  Clause #18 (by clausification #[16]): ∀ (a : Iota), Eq (big_g a) True
% 5.72/5.97  Clause #22 (by superposition #[18, 6]): Eq True False
% 5.72/5.97  Clause #23 (by clausification #[22]): False
% 5.72/5.97  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------