TSTP Solution File: SYN986+1.002 by Duper---1.0

View Problem - Process Solution

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% File     : Duper---1.0
% Problem  : SYN986+1.002 : TPTP v8.1.2. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n032.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 02:13:33 EDT 2023

% Result   : Theorem 3.88s 4.05s
% Output   : Proof 3.88s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.11  % Problem    : SYN986+1.002 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.12  % Command    : duper %s
% 0.11/0.32  % Computer : n032.cluster.edu
% 0.11/0.32  % Model    : x86_64 x86_64
% 0.11/0.32  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.32  % Memory   : 8042.1875MB
% 0.11/0.32  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.11/0.32  % CPULimit   : 300
% 0.11/0.32  % WCLimit    : 300
% 0.11/0.32  % DateTime   : Sat Aug 26 21:02:29 EDT 2023
% 0.11/0.32  % CPUTime    : 
% 3.88/4.05  SZS status Theorem for theBenchmark.p
% 3.88/4.05  SZS output start Proof for theBenchmark.p
% 3.88/4.05  Clause #0 (by assumption #[]): Eq (∀ (Y : Iota), r Y zero (succ Y)) True
% 3.88/4.05  Clause #1 (by assumption #[]): Eq (∀ (Y X Z Z1 : Iota), r Y X Z → r Z X Z1 → r Y (succ X) Z1) True
% 3.88/4.05  Clause #2 (by assumption #[]): Eq (Not (Exists fun Z2 => Exists fun Z1 => Exists fun Z0 => And (And (r zero zero Z2) (r zero Z2 Z1)) (r zero Z1 Z0)))
% 3.88/4.05    True
% 3.88/4.05  Clause #3 (by clausification #[0]): ∀ (a : Iota), Eq (r a zero (succ a)) True
% 3.88/4.05  Clause #4 (by clausification #[1]): ∀ (a : Iota), Eq (∀ (X Z Z1 : Iota), r a X Z → r Z X Z1 → r a (succ X) Z1) True
% 3.88/4.05  Clause #5 (by clausification #[4]): ∀ (a a_1 : Iota), Eq (∀ (Z Z1 : Iota), r a a_1 Z → r Z a_1 Z1 → r a (succ a_1) Z1) True
% 3.88/4.05  Clause #6 (by clausification #[5]): ∀ (a a_1 a_2 : Iota), Eq (∀ (Z1 : Iota), r a a_1 a_2 → r a_2 a_1 Z1 → r a (succ a_1) Z1) True
% 3.88/4.05  Clause #7 (by clausification #[6]): ∀ (a a_1 a_2 a_3 : Iota), Eq (r a a_1 a_2 → r a_2 a_1 a_3 → r a (succ a_1) a_3) True
% 3.88/4.05  Clause #8 (by clausification #[7]): ∀ (a a_1 a_2 a_3 : Iota), Or (Eq (r a a_1 a_2) False) (Eq (r a_2 a_1 a_3 → r a (succ a_1) a_3) True)
% 3.88/4.05  Clause #9 (by clausification #[8]): ∀ (a a_1 a_2 a_3 : Iota), Or (Eq (r a a_1 a_2) False) (Or (Eq (r a_2 a_1 a_3) False) (Eq (r a (succ a_1) a_3) True))
% 3.88/4.05  Clause #10 (by superposition #[9, 3]): ∀ (a a_1 : Iota), Or (Eq (r (succ a) zero a_1) False) (Or (Eq (r a (succ zero) a_1) True) (Eq False True))
% 3.88/4.05  Clause #11 (by clausification #[10]): ∀ (a a_1 : Iota), Or (Eq (r (succ a) zero a_1) False) (Eq (r a (succ zero) a_1) True)
% 3.88/4.05  Clause #12 (by superposition #[11, 3]): ∀ (a : Iota), Or (Eq (r a (succ zero) (succ (succ a))) True) (Eq False True)
% 3.88/4.05  Clause #13 (by clausification #[12]): ∀ (a : Iota), Eq (r a (succ zero) (succ (succ a))) True
% 3.88/4.05  Clause #14 (by superposition #[13, 9]): ∀ (a a_1 : Iota),
% 3.88/4.05    Or (Eq True False) (Or (Eq (r (succ (succ a)) (succ zero) a_1) False) (Eq (r a (succ (succ zero)) a_1) True))
% 3.88/4.05  Clause #15 (by clausification #[14]): ∀ (a a_1 : Iota), Or (Eq (r (succ (succ a)) (succ zero) a_1) False) (Eq (r a (succ (succ zero)) a_1) True)
% 3.88/4.05  Clause #16 (by superposition #[15, 13]): ∀ (a : Iota), Or (Eq (r a (succ (succ zero)) (succ (succ (succ (succ a))))) True) (Eq False True)
% 3.88/4.05  Clause #17 (by clausification #[2]): Eq (Exists fun Z2 => Exists fun Z1 => Exists fun Z0 => And (And (r zero zero Z2) (r zero Z2 Z1)) (r zero Z1 Z0)) False
% 3.88/4.05  Clause #18 (by clausification #[17]): ∀ (a : Iota), Eq (Exists fun Z1 => Exists fun Z0 => And (And (r zero zero a) (r zero a Z1)) (r zero Z1 Z0)) False
% 3.88/4.05  Clause #19 (by clausification #[18]): ∀ (a a_1 : Iota), Eq (Exists fun Z0 => And (And (r zero zero a) (r zero a a_1)) (r zero a_1 Z0)) False
% 3.88/4.05  Clause #20 (by clausification #[19]): ∀ (a a_1 a_2 : Iota), Eq (And (And (r zero zero a) (r zero a a_1)) (r zero a_1 a_2)) False
% 3.88/4.05  Clause #21 (by clausification #[20]): ∀ (a a_1 a_2 : Iota), Or (Eq (And (r zero zero a) (r zero a a_1)) False) (Eq (r zero a_1 a_2) False)
% 3.88/4.05  Clause #22 (by clausification #[21]): ∀ (a a_1 a_2 : Iota), Or (Eq (r zero a a_1) False) (Or (Eq (r zero zero a_2) False) (Eq (r zero a_2 a) False))
% 3.88/4.05  Clause #28 (by clausification #[16]): ∀ (a : Iota), Eq (r a (succ (succ zero)) (succ (succ (succ (succ a))))) True
% 3.88/4.05  Clause #30 (by superposition #[28, 22]): ∀ (a : Iota), Or (Eq True False) (Or (Eq (r zero zero a) False) (Eq (r zero a (succ (succ zero))) False))
% 3.88/4.05  Clause #36 (by clausification #[30]): ∀ (a : Iota), Or (Eq (r zero zero a) False) (Eq (r zero a (succ (succ zero))) False)
% 3.88/4.05  Clause #37 (by superposition #[36, 3]): Or (Eq (r zero (succ zero) (succ (succ zero))) False) (Eq False True)
% 3.88/4.05  Clause #38 (by clausification #[37]): Eq (r zero (succ zero) (succ (succ zero))) False
% 3.88/4.05  Clause #39 (by superposition #[38, 13]): Eq False True
% 3.88/4.05  Clause #40 (by clausification #[39]): False
% 3.88/4.05  SZS output end Proof for theBenchmark.p
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