↑ Up

Duper---1.0.THM-Prf.s

View TPTP
Problem
Process solution in
SystemOnTSTP
Download .tgz
%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : SEV015^5 : TPTP v9.2.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n007.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 08:02:44 PM UTC 2025

% Result   : Theorem 4.35s 4.53s
% Output   : Proof 4.36s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.09/0.14  % Problem    : SEV015^5 : TPTP v9.2.0. Released v4.0.0.
% 0.09/0.15  % Command    : duper %s
% 0.15/0.37  % Computer : n007.cluster.edu
% 0.15/0.37  % Model    : x86_64 x86_64
% 0.15/0.37  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.37  % Memory   : 8042.1875MB
% 0.15/0.37  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.15/0.37  % CPULimit   : 300
% 0.15/0.37  % WCLimit    : 300
% 0.15/0.37  % DateTime   : Fri Oct  3 01:03:23 EDT 2025
% 0.15/0.37  % CPUTime    : 
% 4.35/4.53  SZS status Theorem for theBenchmark.p
% 4.35/4.53  SZS output start Proof for theBenchmark.p
% 4.35/4.53  Clause #0 (by assumption #[]): Eq
% 4.35/4.53    (Not
% 4.35/4.53      (∀ (Xp : a → a → Prop),
% 4.35/4.53        And (And (∀ (Xx Xy : a), Xp Xx Xy → Xp Xy Xx) (∀ (Xx Xy Xz : a), And (Xp Xx Xy) (Xp Xy Xz) → Xp Xx Xz))
% 4.35/4.53            (Eq Xp Xp) →
% 4.35/4.53          ∀ (Xx Xy : a), Xp Xx Xy → Xp Xx Xx))
% 4.35/4.53    True
% 4.35/4.53  Clause #1 (by clausification #[0]): Eq
% 4.35/4.53    (∀ (Xp : a → a → Prop),
% 4.35/4.53      And (And (∀ (Xx Xy : a), Xp Xx Xy → Xp Xy Xx) (∀ (Xx Xy Xz : a), And (Xp Xx Xy) (Xp Xy Xz) → Xp Xx Xz)) (Eq Xp Xp) →
% 4.35/4.53        ∀ (Xx Xy : a), Xp Xx Xy → Xp Xx Xx)
% 4.35/4.53    False
% 4.35/4.53  Clause #2 (by clausification #[1]): ∀ (a_1 : a → a → Prop),
% 4.35/4.53    Eq
% 4.35/4.53      (Not
% 4.35/4.53        (And
% 4.35/4.53            (And (∀ (Xx Xy : a), skS.0 0 a_1 Xx Xy → skS.0 0 a_1 Xy Xx)
% 4.35/4.53              (∀ (Xx Xy Xz : a), And (skS.0 0 a_1 Xx Xy) (skS.0 0 a_1 Xy Xz) → skS.0 0 a_1 Xx Xz))
% 4.35/4.53            (Eq (skS.0 0 a_1) (skS.0 0 a_1)) →
% 4.35/4.53          ∀ (Xx Xy : a), skS.0 0 a_1 Xx Xy → skS.0 0 a_1 Xx Xx))
% 4.35/4.53      True
% 4.35/4.53  Clause #3 (by clausification #[2]): ∀ (a_1 : a → a → Prop),
% 4.35/4.53    Eq
% 4.35/4.53      (And
% 4.35/4.53          (And (∀ (Xx Xy : a), skS.0 0 a_1 Xx Xy → skS.0 0 a_1 Xy Xx)
% 4.35/4.53            (∀ (Xx Xy Xz : a), And (skS.0 0 a_1 Xx Xy) (skS.0 0 a_1 Xy Xz) → skS.0 0 a_1 Xx Xz))
% 4.35/4.53          (Eq (skS.0 0 a_1) (skS.0 0 a_1)) →
% 4.35/4.53        ∀ (Xx Xy : a), skS.0 0 a_1 Xx Xy → skS.0 0 a_1 Xx Xx)
% 4.35/4.53      False
% 4.35/4.53  Clause #4 (by clausification #[3]): ∀ (a_1 : a → a → Prop),
% 4.35/4.53    Eq
% 4.35/4.53      (And
% 4.35/4.53        (And (∀ (Xx Xy : a), skS.0 0 a_1 Xx Xy → skS.0 0 a_1 Xy Xx)
% 4.35/4.53          (∀ (Xx Xy Xz : a), And (skS.0 0 a_1 Xx Xy) (skS.0 0 a_1 Xy Xz) → skS.0 0 a_1 Xx Xz))
% 4.35/4.53        (Eq (skS.0 0 a_1) (skS.0 0 a_1)))
% 4.35/4.53      True
% 4.35/4.53  Clause #5 (by clausification #[3]): ∀ (a_1 : a → a → Prop), Eq (∀ (Xx Xy : a), skS.0 0 a_1 Xx Xy → skS.0 0 a_1 Xx Xx) False
% 4.35/4.53  Clause #7 (by clausification #[4]): ∀ (a_1 : a → a → Prop),
% 4.35/4.53    Eq
% 4.35/4.53      (And (∀ (Xx Xy : a), skS.0 0 a_1 Xx Xy → skS.0 0 a_1 Xy Xx)
% 4.35/4.53        (∀ (Xx Xy Xz : a), And (skS.0 0 a_1 Xx Xy) (skS.0 0 a_1 Xy Xz) → skS.0 0 a_1 Xx Xz))
% 4.35/4.53      True
% 4.35/4.53  Clause #9 (by clausification #[5]): ∀ (a_1 : a → a → Prop) (a_2 : a),
% 4.35/4.53    Eq (Not (∀ (Xy : a), skS.0 0 a_1 (skS.0 1 a_1 a_2) Xy → skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 1 a_1 a_2))) True
% 4.35/4.53  Clause #10 (by clausification #[9]): ∀ (a_1 : a → a → Prop) (a_2 : a),
% 4.35/4.53    Eq (∀ (Xy : a), skS.0 0 a_1 (skS.0 1 a_1 a_2) Xy → skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 1 a_1 a_2)) False
% 4.35/4.53  Clause #11 (by clausification #[10]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a),
% 4.35/4.53    Eq (Not (skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3) → skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 1 a_1 a_2))) True
% 4.35/4.53  Clause #12 (by clausification #[11]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a),
% 4.35/4.53    Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3) → skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 1 a_1 a_2)) False
% 4.35/4.53  Clause #13 (by clausification #[12]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a), Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3)) True
% 4.35/4.53  Clause #14 (by clausification #[12]): ∀ (a_1 : a → a → Prop) (a_2 : a), Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 1 a_1 a_2)) False
% 4.35/4.53  Clause #15 (by clausification #[7]): ∀ (a_1 : a → a → Prop), Eq (∀ (Xx Xy Xz : a), And (skS.0 0 a_1 Xx Xy) (skS.0 0 a_1 Xy Xz) → skS.0 0 a_1 Xx Xz) True
% 4.35/4.53  Clause #16 (by clausification #[7]): ∀ (a_1 : a → a → Prop), Eq (∀ (Xx Xy : a), skS.0 0 a_1 Xx Xy → skS.0 0 a_1 Xy Xx) True
% 4.35/4.53  Clause #17 (by clausification #[15]): ∀ (a_1 : a → a → Prop) (a_2 : a),
% 4.35/4.53    Eq (∀ (Xy Xz : a), And (skS.0 0 a_1 a_2 Xy) (skS.0 0 a_1 Xy Xz) → skS.0 0 a_1 a_2 Xz) True
% 4.35/4.53  Clause #18 (by clausification #[17]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a),
% 4.35/4.53    Eq (∀ (Xz : a), And (skS.0 0 a_1 a_2 a_3) (skS.0 0 a_1 a_3 Xz) → skS.0 0 a_1 a_2 Xz) True
% 4.35/4.53  Clause #19 (by clausification #[18]): ∀ (a_1 : a → a → Prop) (a_2 a_3 a_4 : a),
% 4.35/4.53    Eq (And (skS.0 0 a_1 a_2 a_3) (skS.0 0 a_1 a_3 a_4) → skS.0 0 a_1 a_2 a_4) True
% 4.35/4.53  Clause #20 (by clausification #[19]): ∀ (a_1 : a → a → Prop) (a_2 a_3 a_4 : a),
% 4.35/4.53    Or (Eq (And (skS.0 0 a_1 a_2 a_3) (skS.0 0 a_1 a_3 a_4)) False) (Eq (skS.0 0 a_1 a_2 a_4) True)
% 4.36/4.55  Clause #21 (by clausification #[20]): ∀ (a_1 : a → a → Prop) (a_2 a_3 a_4 : a),
% 4.36/4.55    Or (Eq (skS.0 0 a_1 a_2 a_3) True) (Or (Eq (skS.0 0 a_1 a_2 a_4) False) (Eq (skS.0 0 a_1 a_4 a_3) False))
% 4.36/4.55  Clause #22 (by superposition #[21, 13]): ∀ (a_1 : a → a → Prop) (a_2 a_3 a_4 : a),
% 4.36/4.55    Or (Eq (skS.0 0 (fun x x_1 => a_1 x x_1) (skS.0 1 a_1 a_2) a_3) True)
% 4.36/4.55      (Or (Eq (skS.0 0 (fun x x_1 => a_1 x x_1) (skS.0 2 a_1 a_2 a_4) a_3) False) (Eq False True))
% 4.36/4.55  Clause #23 (by clausification #[16]): ∀ (a_1 : a → a → Prop) (a_2 : a), Eq (∀ (Xy : a), skS.0 0 a_1 a_2 Xy → skS.0 0 a_1 Xy a_2) True
% 4.36/4.55  Clause #24 (by clausification #[23]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a), Eq (skS.0 0 a_1 a_2 a_3 → skS.0 0 a_1 a_3 a_2) True
% 4.36/4.55  Clause #25 (by clausification #[24]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a), Or (Eq (skS.0 0 a_1 a_2 a_3) False) (Eq (skS.0 0 a_1 a_3 a_2) True)
% 4.36/4.55  Clause #26 (by superposition #[25, 13]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a),
% 4.36/4.55    Or (Eq (skS.0 0 (fun x x_1 => a_1 x x_1) (skS.0 2 a_1 a_2 a_3) (skS.0 1 a_1 a_2)) True) (Eq False True)
% 4.36/4.55  Clause #27 (by betaEtaReduce #[22]): ∀ (a_1 : a → a → Prop) (a_2 a_3 a_4 : a),
% 4.36/4.55    Or (Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) a_3) True)
% 4.36/4.55      (Or (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_4) a_3) False) (Eq False True))
% 4.36/4.55  Clause #28 (by clausification #[27]): ∀ (a_1 : a → a → Prop) (a_2 a_3 a_4 : a),
% 4.36/4.55    Or (Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) a_3) True) (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_4) a_3) False)
% 4.36/4.55  Clause #29 (by betaEtaReduce #[26]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a), Or (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3) (skS.0 1 a_1 a_2)) True) (Eq False True)
% 4.36/4.55  Clause #30 (by clausification #[29]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3) (skS.0 1 a_1 a_2)) True
% 4.36/4.55  Clause #31 (by superposition #[30, 28]): ∀ (a_1 : a → a → Prop) (a_2 : a), Or (Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 1 a_1 a_2)) True) (Eq True False)
% 4.36/4.55  Clause #34 (by clausification #[31]): ∀ (a_1 : a → a → Prop) (a_2 : a), Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 1 a_1 a_2)) True
% 4.36/4.55  Clause #35 (by superposition #[34, 14]): Eq True False
% 4.36/4.55  Clause #37 (by clausification #[35]): False
% 4.36/4.55  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------