TSTP Solution File: LCL773_5 by Duper---1.0

View Problem - Process Solution

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% File     : Duper---1.0
% Problem  : LCL773_5 : TPTP v8.1.2. Released v6.0.0.
% 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 : Thu Aug 31 07:11:10 EDT 2023

% Result   : Theorem 7.05s 7.25s
% Output   : Proof 7.05s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.10  % Problem    : LCL773_5 : TPTP v8.1.2. Released v6.0.0.
% 0.00/0.11  % Command    : duper %s
% 0.12/0.31  % Computer : n032.cluster.edu
% 0.12/0.31  % Model    : x86_64 x86_64
% 0.12/0.31  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.31  % Memory   : 8042.1875MB
% 0.12/0.31  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.31  % CPULimit   : 300
% 0.12/0.31  % WCLimit    : 300
% 0.12/0.31  % DateTime   : Fri Aug 25 05:23:42 EDT 2023
% 0.12/0.31  % CPUTime    : 
% 7.05/7.25  SZS status Theorem for theBenchmark.p
% 7.05/7.25  SZS output start Proof for theBenchmark.p
% 7.05/7.25  Clause #1 (by assumption #[]): Eq (∀ (J I : nat) (R2 : dB), pp (aa dB bool it R2) → pp (aa dB bool it (subst R2 (var I) J))) True
% 7.05/7.25  Clause #109 (by assumption #[]): Eq (pp (aa dB bool it r)) True
% 7.05/7.25  Clause #111 (by assumption #[]): Eq (Not (pp (aa dB bool it (subst r (var i) (zero_zero nat))))) True
% 7.05/7.25  Clause #113 (by clausification #[1]): ∀ (a : nat), Eq (∀ (I : nat) (R2 : dB), pp (aa dB bool it R2) → pp (aa dB bool it (subst R2 (var I) a))) True
% 7.05/7.25  Clause #114 (by clausification #[113]): ∀ (a a_1 : nat), Eq (∀ (R2 : dB), pp (aa dB bool it R2) → pp (aa dB bool it (subst R2 (var a) a_1))) True
% 7.05/7.25  Clause #115 (by clausification #[114]): ∀ (a : dB) (a_1 a_2 : nat), Eq (pp (aa dB bool it a) → pp (aa dB bool it (subst a (var a_1) a_2))) True
% 7.05/7.25  Clause #116 (by clausification #[115]): ∀ (a : dB) (a_1 a_2 : nat), Or (Eq (pp (aa dB bool it a)) False) (Eq (pp (aa dB bool it (subst a (var a_1) a_2))) True)
% 7.05/7.25  Clause #119 (by superposition #[109, 116]): ∀ (a a_1 : nat), Or (Eq (pp (aa dB bool it (subst r (var a) a_1))) True) (Eq False True)
% 7.05/7.25  Clause #2273 (by clausification #[111]): Eq (pp (aa dB bool it (subst r (var i) (zero_zero nat)))) False
% 7.05/7.25  Clause #2357 (by clausification #[119]): ∀ (a a_1 : nat), Eq (pp (aa dB bool it (subst r (var a) a_1))) True
% 7.05/7.25  Clause #2361 (by superposition #[2357, 2273]): Eq True False
% 7.05/7.25  Clause #2392 (by clausification #[2361]): False
% 7.05/7.25  SZS output end Proof for theBenchmark.p
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