TSTP Solution File: SYN983^1 by Duper---1.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SYN983^1 : TPTP v8.1.2. Released v3.6.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n003.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 02:13:32 EDT 2023

% Result   : Theorem 3.67s 3.84s
% Output   : Proof 3.67s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem    : SYN983^1 : TPTP v8.1.2. Released v3.6.0.
% 0.13/0.14  % Command    : duper %s
% 0.13/0.35  % Computer : n003.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   : Sat Aug 26 20:25:09 EDT 2023
% 0.20/0.35  % CPUTime    : 
% 3.67/3.84  SZS status Theorem for theBenchmark.p
% 3.67/3.84  SZS output start Proof for theBenchmark.p
% 3.67/3.84  Clause #0 (by assumption #[]): Eq (Not (And (p a) (p b) → p (And a b))) True
% 3.67/3.84  Clause #1 (by clausification #[0]): Eq (And (p a) (p b) → p (And a b)) False
% 3.67/3.84  Clause #2 (by clausification #[1]): Eq (And (p a) (p b)) True
% 3.67/3.84  Clause #3 (by clausification #[1]): Eq (p (And a b)) False
% 3.67/3.84  Clause #4 (by clausification #[2]): Eq (p b) True
% 3.67/3.84  Clause #5 (by clausification #[2]): Eq (p a) True
% 3.67/3.84  Clause #7 (by identity boolHoist #[4]): Or (Eq (p False) True) (Eq b True)
% 3.67/3.84  Clause #8 (by identity loobHoist #[5]): Or (Eq (p True) True) (Eq a False)
% 3.67/3.84  Clause #9 (by identity boolHoist #[5]): Or (Eq (p False) True) (Eq a True)
% 3.67/3.84  Clause #10 (by identity loobHoist #[3]): Or (Eq (p True) False) (Eq (And a b) False)
% 3.67/3.84  Clause #11 (by identity boolHoist #[3]): Or (Eq (p False) False) (Eq (And a b) True)
% 3.67/3.84  Clause #12 (by clausification #[10]): Or (Eq (p True) False) (Or (Eq a False) (Eq b False))
% 3.67/3.84  Clause #13 (by clausification #[11]): Or (Eq (p False) False) (Eq b True)
% 3.67/3.84  Clause #14 (by clausification #[11]): Or (Eq (p False) False) (Eq a True)
% 3.67/3.84  Clause #15 (by superposition #[13, 7]): Or (Eq b True) (Or (Eq False True) (Eq b True))
% 3.67/3.84  Clause #18 (by superposition #[14, 9]): Or (Eq a True) (Or (Eq False True) (Eq a True))
% 3.67/3.84  Clause #19 (by clausification #[18]): Or (Eq a True) (Eq a True)
% 3.67/3.84  Clause #20 (by eliminate duplicate literals #[19]): Eq a True
% 3.67/3.84  Clause #21 (by backward demodulation #[20, 8]): Or (Eq (p True) True) (Eq True False)
% 3.67/3.84  Clause #22 (by backward demodulation #[20, 12]): Or (Eq (p True) False) (Or (Eq True False) (Eq b False))
% 3.67/3.84  Clause #25 (by clausification #[21]): Eq (p True) True
% 3.67/3.84  Clause #26 (by clausification #[15]): Or (Eq b True) (Eq b True)
% 3.67/3.84  Clause #27 (by eliminate duplicate literals #[26]): Eq b True
% 3.67/3.84  Clause #32 (by clausification #[22]): Or (Eq (p True) False) (Eq b False)
% 3.67/3.84  Clause #33 (by forward demodulation #[32, 27]): Or (Eq (p True) False) (Eq True False)
% 3.67/3.84  Clause #34 (by clausification #[33]): Eq (p True) False
% 3.67/3.84  Clause #35 (by superposition #[34, 25]): Eq False True
% 3.67/3.84  Clause #36 (by clausification #[35]): False
% 3.67/3.84  SZS output end Proof for theBenchmark.p
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