TSTP Solution File: SYN367+1 by Duper---1.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SYN367+1 : TPTP v8.1.2. Released v2.0.0.
% 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 Sep  1 02:11:27 EDT 2023

% Result   : Theorem 3.35s 3.56s
% Output   : Proof 3.35s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem    : SYN367+1 : TPTP v8.1.2. Released v2.0.0.
% 0.00/0.13  % Command    : duper %s
% 0.13/0.34  % Computer : n008.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   : Sat Aug 26 17:36:47 EDT 2023
% 0.13/0.35  % CPUTime    : 
% 3.35/3.56  SZS status Theorem for theBenchmark.p
% 3.35/3.56  SZS output start Proof for theBenchmark.p
% 3.35/3.56  Clause #0 (by assumption #[]): Eq
% 3.35/3.56    (Not
% 3.35/3.56      ((∀ (X : Iota), Or (And p (big_q X)) (And (Not p) (big_r X))) → Or (∀ (X : Iota), big_q X) (∀ (X : Iota), big_r X)))
% 3.35/3.56    True
% 3.35/3.56  Clause #1 (by clausification #[0]): Eq ((∀ (X : Iota), Or (And p (big_q X)) (And (Not p) (big_r X))) → Or (∀ (X : Iota), big_q X) (∀ (X : Iota), big_r X))
% 3.35/3.56    False
% 3.35/3.56  Clause #2 (by clausification #[1]): Eq (∀ (X : Iota), Or (And p (big_q X)) (And (Not p) (big_r X))) True
% 3.35/3.56  Clause #3 (by clausification #[1]): Eq (Or (∀ (X : Iota), big_q X) (∀ (X : Iota), big_r X)) False
% 3.35/3.56  Clause #4 (by clausification #[2]): ∀ (a : Iota), Eq (Or (And p (big_q a)) (And (Not p) (big_r a))) True
% 3.35/3.56  Clause #5 (by clausification #[4]): ∀ (a : Iota), Or (Eq (And p (big_q a)) True) (Eq (And (Not p) (big_r a)) True)
% 3.35/3.56  Clause #6 (by clausification #[5]): ∀ (a : Iota), Or (Eq (And (Not p) (big_r a)) True) (Eq (big_q a) True)
% 3.35/3.56  Clause #7 (by clausification #[5]): ∀ (a : Iota), Or (Eq (And (Not p) (big_r a)) True) (Eq p True)
% 3.35/3.56  Clause #9 (by clausification #[6]): ∀ (a : Iota), Or (Eq (big_q a) True) (Eq (Not p) True)
% 3.35/3.56  Clause #10 (by clausification #[9]): ∀ (a : Iota), Or (Eq (big_q a) True) (Eq p False)
% 3.35/3.56  Clause #11 (by clausification #[7]): ∀ (a : Iota), Or (Eq p True) (Eq (big_r a) True)
% 3.35/3.56  Clause #14 (by clausification #[3]): Eq (∀ (X : Iota), big_r X) False
% 3.35/3.56  Clause #15 (by clausification #[3]): Eq (∀ (X : Iota), big_q X) False
% 3.35/3.56  Clause #16 (by clausification #[14]): ∀ (a : Iota), Eq (Not (big_r (skS.0 0 a))) True
% 3.35/3.56  Clause #17 (by clausification #[16]): ∀ (a : Iota), Eq (big_r (skS.0 0 a)) False
% 3.35/3.56  Clause #19 (by superposition #[17, 11]): Or (Eq p True) (Eq False True)
% 3.35/3.56  Clause #20 (by clausification #[15]): ∀ (a : Iota), Eq (Not (big_q (skS.0 1 a))) True
% 3.35/3.56  Clause #21 (by clausification #[20]): ∀ (a : Iota), Eq (big_q (skS.0 1 a)) False
% 3.35/3.56  Clause #22 (by clausification #[19]): Eq p True
% 3.35/3.56  Clause #23 (by backward demodulation #[22, 10]): ∀ (a : Iota), Or (Eq (big_q a) True) (Eq True False)
% 3.35/3.56  Clause #25 (by clausification #[23]): ∀ (a : Iota), Eq (big_q a) True
% 3.35/3.56  Clause #26 (by superposition #[25, 21]): Eq True False
% 3.35/3.56  Clause #27 (by clausification #[26]): False
% 3.35/3.56  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------