TSTP Solution File: SWW469_1 by Duper---1.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SWW469_1 : TPTP v8.1.2. Released v5.3.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n013.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 00:26:13 EDT 2023

% Result   : Theorem 3.45s 3.67s
% Output   : Proof 3.45s
% 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    : SWW469_1 : TPTP v8.1.2. Released v5.3.0.
% 0.00/0.13  % Command    : duper %s
% 0.12/0.34  % Computer : n013.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34  % CPULimit   : 300
% 0.12/0.34  % WCLimit    : 300
% 0.12/0.34  % DateTime   : Sun Aug 27 21:21:17 EDT 2023
% 0.19/0.35  % CPUTime    : 
% 3.45/3.67  SZS status Theorem for theBenchmark.p
% 3.45/3.67  SZS output start Proof for theBenchmark.p
% 3.45/3.67  Clause #0 (by assumption #[]): Eq (Iff hoare_1310879719gleton (Exists fun S => Exists fun T => Ne S T)) True
% 3.45/3.67  Clause #3 (by assumption #[]): Eq hoare_1310879719gleton True
% 3.45/3.67  Clause #4 (by assumption #[]): Eq (Not (∀ (T : state), Not (∀ (S : state), Eq S T))) True
% 3.45/3.67  Clause #6 (by clausification #[4]): Eq (∀ (T : state), Not (∀ (S : state), Eq S T)) False
% 3.45/3.67  Clause #7 (by clausification #[6]): ∀ (a : state), Eq (Not (Not (∀ (S : state), Eq S (skS.0 0 a)))) True
% 3.45/3.67  Clause #8 (by clausification #[7]): ∀ (a : state), Eq (Not (∀ (S : state), Eq S (skS.0 0 a))) False
% 3.45/3.67  Clause #9 (by clausification #[8]): ∀ (a : state), Eq (∀ (S : state), Eq S (skS.0 0 a)) True
% 3.45/3.67  Clause #10 (by clausification #[9]): ∀ (a a_1 : state), Eq (Eq a (skS.0 0 a_1)) True
% 3.45/3.67  Clause #11 (by clausification #[10]): ∀ (a a_1 : state), Eq a (skS.0 0 a_1)
% 3.45/3.67  Clause #12 (by superposition #[11, 11]): ∀ (a a_1 : state), Eq a a_1
% 3.45/3.67  Clause #13 (by betaEtaReduce #[0]): Eq (Iff hoare_1310879719gleton (Exists fun S => Exists (Ne S))) True
% 3.45/3.67  Clause #15 (by clausification #[13]): Or (Eq hoare_1310879719gleton False) (Eq (Exists fun S => Exists (Ne S)) True)
% 3.45/3.67  Clause #19 (by clausification #[15]): ∀ (a : state), Or (Eq hoare_1310879719gleton False) (Eq (Exists (Ne (skS.0 1 a))) True)
% 3.45/3.67  Clause #20 (by clausification #[19]): ∀ (a a_1 : state), Or (Eq hoare_1310879719gleton False) (Eq (Ne (skS.0 1 a) (skS.0 2 a a_1)) True)
% 3.45/3.67  Clause #21 (by clausification #[20]): ∀ (a a_1 : state), Or (Eq hoare_1310879719gleton False) (Ne (skS.0 1 a) (skS.0 2 a a_1))
% 3.45/3.67  Clause #22 (by forward demodulation #[21, 3]): ∀ (a a_1 : state), Or (Eq True False) (Ne (skS.0 1 a) (skS.0 2 a a_1))
% 3.45/3.67  Clause #23 (by clausification #[22]): ∀ (a a_1 : state), Ne (skS.0 1 a) (skS.0 2 a a_1)
% 3.45/3.67  Clause #24 (by forward contextual literal cutting #[23, 12]): False
% 3.45/3.67  SZS output end Proof for theBenchmark.p
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