TSTP Solution File: SYN001+1 by Duper---1.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SYN001+1 : TPTP v8.1.2. Released v2.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n006.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 02:10:07 EDT 2023

% Result   : Theorem 3.65s 3.84s
% Output   : Proof 3.65s
% 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    : SYN001+1 : TPTP v8.1.2. Released v2.0.0.
% 0.00/0.13  % Command    : duper %s
% 0.13/0.34  % Computer : n006.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 17:01:37 EDT 2023
% 0.13/0.34  % CPUTime    : 
% 3.65/3.84  SZS status Theorem for theBenchmark.p
% 3.65/3.84  SZS output start Proof for theBenchmark.p
% 3.65/3.84  Clause #0 (by assumption #[]): Eq (Not (Iff (Not (Not p)) p)) True
% 3.65/3.84  Clause #1 (by clausification #[0]): Eq (Iff (Not (Not p)) p) False
% 3.65/3.84  Clause #2 (by clausification #[1]): Or (Eq (Not (Not p)) False) (Eq p False)
% 3.65/3.84  Clause #3 (by clausification #[1]): Or (Eq (Not (Not p)) True) (Eq p True)
% 3.65/3.84  Clause #4 (by clausification #[2]): Or (Eq p False) (Eq (Not p) True)
% 3.65/3.84  Clause #5 (by clausification #[4]): Or (Eq p False) (Eq p False)
% 3.65/3.84  Clause #6 (by eliminate duplicate literals #[5]): Eq p False
% 3.65/3.84  Clause #7 (by clausification #[3]): Or (Eq p True) (Eq (Not p) False)
% 3.65/3.84  Clause #8 (by clausification #[7]): Or (Eq p True) (Eq p True)
% 3.65/3.84  Clause #9 (by eliminate duplicate literals #[8]): Eq p True
% 3.65/3.84  Clause #10 (by superposition #[9, 6]): Eq True False
% 3.65/3.84  Clause #11 (by clausification #[10]): False
% 3.65/3.84  SZS output end Proof for theBenchmark.p
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