%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : ITP209^2 : TPTP v8.1.2. Released v8.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n014.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 03:39:19 EDT 2023

% Result   : Theorem 5.59s 5.80s
% Output   : Proof 5.59s
% 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    : ITP209^2 : TPTP v8.1.2. Released v8.0.0.
% 0.00/0.13  % Command    : duper %s
% 0.14/0.34  % Computer : n014.cluster.edu
% 0.14/0.34  % Model    : x86_64 x86_64
% 0.14/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34  % Memory   : 8042.1875MB
% 0.14/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34  % CPULimit   : 300
% 0.14/0.34  % WCLimit    : 300
% 0.14/0.34  % DateTime   : Sun Aug 27 15:58:16 EDT 2023
% 0.14/0.35  % CPUTime    : 
% 5.59/5.80  SZS status Theorem for theBenchmark.p
% 5.59/5.80  SZS output start Proof for theBenchmark.p
% 5.59/5.80  Clause #1 (by assumption #[]): Eq (∀ (A2 B2 : a), Eq (f A2 B2) (f B2 A2)) True
% 5.59/5.80  Clause #2 (by assumption #[]): Eq (∀ (A2 B2 C : a), Eq (f A2 (f B2 C)) (f B2 (f A2 C))) True
% 5.59/5.80  Clause #12 (by assumption #[]): Eq (Not (Eq (f (f a2 b) c) (f (f a2 c) b))) True
% 5.59/5.80  Clause #50 (by clausification #[1]): ∀ (a_1 : a), Eq (∀ (B2 : a), Eq (f a_1 B2) (f B2 a_1)) True
% 5.59/5.80  Clause #51 (by clausification #[50]): ∀ (a_1 a_2 : a), Eq (Eq (f a_1 a_2) (f a_2 a_1)) True
% 5.59/5.80  Clause #52 (by clausification #[51]): ∀ (a_1 a_2 : a), Eq (f a_1 a_2) (f a_2 a_1)
% 5.59/5.80  Clause #130 (by clausification #[2]): ∀ (a_1 : a), Eq (∀ (B2 C : a), Eq (f a_1 (f B2 C)) (f B2 (f a_1 C))) True
% 5.59/5.80  Clause #131 (by clausification #[130]): ∀ (a_1 a_2 : a), Eq (∀ (C : a), Eq (f a_1 (f a_2 C)) (f a_2 (f a_1 C))) True
% 5.59/5.80  Clause #132 (by clausification #[131]): ∀ (a_1 a_2 a_3 : a), Eq (Eq (f a_1 (f a_2 a_3)) (f a_2 (f a_1 a_3))) True
% 5.59/5.80  Clause #133 (by clausification #[132]): ∀ (a_1 a_2 a_3 : a), Eq (f a_1 (f a_2 a_3)) (f a_2 (f a_1 a_3))
% 5.59/5.80  Clause #139 (by superposition #[133, 52]): ∀ (a_1 a_2 a_3 : a), Eq (f a_1 (f a_2 a_3)) (f a_2 (f a_3 a_1))
% 5.59/5.80  Clause #177 (by superposition #[139, 52]): ∀ (a_1 a_2 a_3 : a), Eq (f (f a_1 a_2) a_3) (f a_2 (f a_3 a_1))
% 5.59/5.80  Clause #1572 (by clausification #[12]): Eq (Eq (f (f a2 b) c) (f (f a2 c) b)) False
% 5.59/5.80  Clause #1573 (by clausification #[1572]): Ne (f (f a2 b) c) (f (f a2 c) b)
% 5.59/5.80  Clause #1574 (by forward demodulation #[1573, 177]): Ne (f b (f c a2)) (f (f a2 c) b)
% 5.59/5.80  Clause #1575 (by forward demodulation #[1574, 133]): Ne (f c (f b a2)) (f (f a2 c) b)
% 5.59/5.80  Clause #1576 (by forward demodulation #[1575, 177]): Ne (f c (f b a2)) (f c (f b a2))
% 5.59/5.80  Clause #1577 (by eliminate resolved literals #[1576]): False
% 5.59/5.80  SZS output end Proof for theBenchmark.p
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