TSTP Solution File: NUM721^1 by Duper---1.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Duper---1.0
% Problem  : NUM721^1 : TPTP v8.1.2. Released v3.7.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : duper %s

% Computer : n017.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 10:57:20 EDT 2023

% Result   : Theorem 5.01s 5.24s
% Output   : Proof 5.01s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.10  % Problem    : NUM721^1 : TPTP v8.1.2. Released v3.7.0.
% 0.00/0.11  % Command    : duper %s
% 0.10/0.31  % Computer : n017.cluster.edu
% 0.10/0.31  % Model    : x86_64 x86_64
% 0.10/0.31  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.10/0.31  % Memory   : 8042.1875MB
% 0.10/0.31  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.10/0.31  % CPULimit   : 300
% 0.10/0.31  % WCLimit    : 300
% 0.10/0.31  % DateTime   : Fri Aug 25 16:49:25 EDT 2023
% 0.10/0.31  % CPUTime    : 
% 5.01/5.24  SZS status Theorem for theBenchmark.p
% 5.01/5.24  SZS output start Proof for theBenchmark.p
% 5.01/5.24  Clause #0 (by assumption #[]): Eq (some fun Xv => diffprop y x Xv) True
% 5.01/5.24  Clause #1 (by assumption #[]): Eq (some fun Xv => diffprop u z Xv) True
% 5.01/5.24  Clause #2 (by assumption #[]): Eq
% 5.01/5.24    (∀ (Xx Xy Xz Xu : nat),
% 5.01/5.24      (some fun Xu => diffprop Xx Xy Xu) →
% 5.01/5.24        (some fun Xu_0 => diffprop Xz Xu Xu_0) → some fun Xu_0 => diffprop (ts Xx Xz) (ts Xy Xu) Xu_0)
% 5.01/5.24    True
% 5.01/5.24  Clause #3 (by assumption #[]): Eq (Not (some fun Xv => diffprop (ts y u) (ts x z) Xv)) True
% 5.01/5.24  Clause #4 (by betaEtaReduce #[1]): Eq (some (diffprop u z)) True
% 5.01/5.24  Clause #5 (by betaEtaReduce #[0]): Eq (some (diffprop y x)) True
% 5.01/5.24  Clause #6 (by betaEtaReduce #[3]): Eq (Not (some (diffprop (ts y u) (ts x z)))) True
% 5.01/5.24  Clause #7 (by clausification #[6]): Eq (some (diffprop (ts y u) (ts x z))) False
% 5.01/5.24  Clause #8 (by betaEtaReduce #[2]): Eq (∀ (Xx Xy Xz Xu : nat), some (diffprop Xx Xy) → some (diffprop Xz Xu) → some (diffprop (ts Xx Xz) (ts Xy Xu))) True
% 5.01/5.24  Clause #9 (by clausification #[8]): ∀ (a : nat),
% 5.01/5.24    Eq (∀ (Xy Xz Xu : nat), some (diffprop a Xy) → some (diffprop Xz Xu) → some (diffprop (ts a Xz) (ts Xy Xu))) True
% 5.01/5.24  Clause #10 (by clausification #[9]): ∀ (a a_1 : nat),
% 5.01/5.24    Eq (∀ (Xz Xu : nat), some (diffprop a a_1) → some (diffprop Xz Xu) → some (diffprop (ts a Xz) (ts a_1 Xu))) True
% 5.01/5.24  Clause #11 (by clausification #[10]): ∀ (a a_1 a_2 : nat),
% 5.01/5.24    Eq (∀ (Xu : nat), some (diffprop a a_1) → some (diffprop a_2 Xu) → some (diffprop (ts a a_2) (ts a_1 Xu))) True
% 5.01/5.24  Clause #12 (by clausification #[11]): ∀ (a a_1 a_2 a_3 : nat),
% 5.01/5.24    Eq (some (diffprop a a_1) → some (diffprop a_2 a_3) → some (diffprop (ts a a_2) (ts a_1 a_3))) True
% 5.01/5.24  Clause #13 (by clausification #[12]): ∀ (a a_1 a_2 a_3 : nat),
% 5.01/5.24    Or (Eq (some (diffprop a a_1)) False) (Eq (some (diffprop a_2 a_3) → some (diffprop (ts a a_2) (ts a_1 a_3))) True)
% 5.01/5.24  Clause #14 (by clausification #[13]): ∀ (a a_1 a_2 a_3 : nat),
% 5.01/5.24    Or (Eq (some (diffprop a a_1)) False)
% 5.01/5.24      (Or (Eq (some (diffprop a_2 a_3)) False) (Eq (some (diffprop (ts a a_2) (ts a_1 a_3))) True))
% 5.01/5.24  Clause #16 (by superposition #[14, 5]): ∀ (a a_1 : nat),
% 5.01/5.24    Or (Eq (some (diffprop a a_1)) False) (Or (Eq (some (diffprop (ts y a) (ts x a_1))) True) (Eq False True))
% 5.01/5.24  Clause #17 (by clausification #[16]): ∀ (a a_1 : nat), Or (Eq (some (diffprop a a_1)) False) (Eq (some (diffprop (ts y a) (ts x a_1))) True)
% 5.01/5.24  Clause #18 (by superposition #[17, 4]): Or (Eq (some (diffprop (ts y u) (ts x z))) True) (Eq False True)
% 5.01/5.24  Clause #27 (by clausification #[18]): Eq (some (diffprop (ts y u) (ts x z))) True
% 5.01/5.24  Clause #28 (by superposition #[27, 7]): Eq True False
% 5.01/5.24  Clause #32 (by clausification #[28]): False
% 5.01/5.24  SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------