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Duper---1.0.THM-Prf.s

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%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SEU841^5 : TPTP v9.2.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n019.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 Oct  3 08:02:32 PM UTC 2025

% Result   : Theorem 3.65s 3.87s
% Output   : Proof 3.65s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.09/0.11  % Problem    : SEU841^5 : TPTP v9.2.0. Released v4.0.0.
% 0.09/0.13  % Command    : duper %s
% 0.12/0.33  % Computer : n019.cluster.edu
% 0.12/0.33  % Model    : x86_64 x86_64
% 0.12/0.33  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33  % Memory   : 8042.1875MB
% 0.12/0.33  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33  % CPULimit   : 300
% 0.12/0.33  % WCLimit    : 300
% 0.12/0.33  % DateTime   : Fri Oct  3 04:38:08 EDT 2025
% 0.12/0.34  % CPUTime    : 
% 3.65/3.87  SZS status Theorem for theBenchmark.p
% 3.65/3.87  SZS output start Proof for theBenchmark.p
% 3.65/3.87  Clause #0 (by assumption #[]): Eq (Not (∀ (S T U : a → Prop), And (Eq S T) (Eq T U) → Eq S U)) True
% 3.65/3.87  Clause #1 (by clausification #[0]): Eq (∀ (S T U : a → Prop), And (Eq S T) (Eq T U) → Eq S U) False
% 3.65/3.87  Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop), Eq (Not (∀ (T U : a → Prop), And (Eq (skS.0 0 a_1) T) (Eq T U) → Eq (skS.0 0 a_1) U)) True
% 3.65/3.87  Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop), Eq (∀ (T U : a → Prop), And (Eq (skS.0 0 a_1) T) (Eq T U) → Eq (skS.0 0 a_1) U) False
% 3.65/3.87  Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 3.65/3.87    Eq (Not (∀ (U : a → Prop), And (Eq (skS.0 0 a_1) (skS.0 1 a_1 a_2)) (Eq (skS.0 1 a_1 a_2) U) → Eq (skS.0 0 a_1) U))
% 3.65/3.87      True
% 3.65/3.87  Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 3.65/3.87    Eq (∀ (U : a → Prop), And (Eq (skS.0 0 a_1) (skS.0 1 a_1 a_2)) (Eq (skS.0 1 a_1 a_2) U) → Eq (skS.0 0 a_1) U) False
% 3.65/3.87  Clause #6 (by clausification #[5]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.65/3.87    Eq
% 3.65/3.87      (Not
% 3.65/3.87        (And (Eq (skS.0 0 a_1) (skS.0 1 a_1 a_2)) (Eq (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3)) →
% 3.65/3.87          Eq (skS.0 0 a_1) (skS.0 2 a_1 a_2 a_3)))
% 3.65/3.87      True
% 3.65/3.87  Clause #7 (by clausification #[6]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.65/3.87    Eq
% 3.65/3.87      (And (Eq (skS.0 0 a_1) (skS.0 1 a_1 a_2)) (Eq (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3)) →
% 3.65/3.87        Eq (skS.0 0 a_1) (skS.0 2 a_1 a_2 a_3))
% 3.65/3.87      False
% 3.65/3.87  Clause #8 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.65/3.87    Eq (And (Eq (skS.0 0 a_1) (skS.0 1 a_1 a_2)) (Eq (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3))) True
% 3.65/3.87  Clause #9 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop), Eq (Eq (skS.0 0 a_1) (skS.0 2 a_1 a_2 a_3)) False
% 3.65/3.87  Clause #10 (by clausification #[8]): ∀ (a_1 a_2 a_3 : a → Prop), Eq (Eq (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3)) True
% 3.65/3.87  Clause #11 (by clausification #[8]): ∀ (a_1 a_2 : a → Prop), Eq (Eq (skS.0 0 a_1) (skS.0 1 a_1 a_2)) True
% 3.65/3.87  Clause #12 (by clausification #[10]): ∀ (a_1 a_2 a_3 : a → Prop), Eq (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3)
% 3.65/3.87  Clause #14 (by clausification #[11]): ∀ (a_1 a_2 : a → Prop), Eq (skS.0 0 a_1) (skS.0 1 a_1 a_2)
% 3.65/3.87  Clause #15 (by backward demodulation #[14, 12]): ∀ (a_1 a_2 a_3 : a → Prop), Eq (skS.0 0 a_1) (skS.0 2 a_1 a_2 a_3)
% 3.65/3.87  Clause #22 (by clausification #[9]): ∀ (a_1 a_2 a_3 : a → Prop), Ne (skS.0 0 a_1) (skS.0 2 a_1 a_2 a_3)
% 3.65/3.87  Clause #23 (by forward demodulation #[22, 15]): ∀ (a_1 : a → Prop), Ne (skS.0 0 a_1) (skS.0 0 a_1)
% 3.65/3.87  Clause #24 (by eliminate resolved literals #[23]): False
% 3.65/3.87  SZS output end Proof for theBenchmark.p
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