TSTP Solution File: SEV161^5 by Duper---1.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SEV161^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n020.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 19:24:23 EDT 2023
% Result : Theorem 3.72s 3.91s
% Output : Proof 3.72s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SEV161^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14 % Command : duper %s
% 0.13/0.35 % Computer : n020.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Thu Aug 24 02:37:16 EDT 2023
% 0.13/0.35 % CPUTime :
% 3.72/3.91 SZS status Theorem for theBenchmark.p
% 3.72/3.91 SZS output start Proof for theBenchmark.p
% 3.72/3.91 Clause #0 (by assumption #[]): Eq (Not (∀ (Xr : a → a → Prop) (Xx Xy : a), Iff (Xr Xx Xy) (Xr Xx Xy))) True
% 3.72/3.91 Clause #1 (by clausification #[0]): Eq (∀ (Xr : a → a → Prop) (Xx Xy : a), Iff (Xr Xx Xy) (Xr Xx Xy)) False
% 3.72/3.91 Clause #2 (by clausification #[1]): ∀ (a_1 : a → a → Prop), Eq (Not (∀ (Xx Xy : a), Iff (skS.0 0 a_1 Xx Xy) (skS.0 0 a_1 Xx Xy))) True
% 3.72/3.91 Clause #3 (by clausification #[2]): ∀ (a_1 : a → a → Prop), Eq (∀ (Xx Xy : a), Iff (skS.0 0 a_1 Xx Xy) (skS.0 0 a_1 Xx Xy)) False
% 3.72/3.91 Clause #4 (by clausification #[3]): ∀ (a_1 : a → a → Prop) (a_2 : a),
% 3.72/3.91 Eq (Not (∀ (Xy : a), Iff (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) Xy))) True
% 3.72/3.91 Clause #5 (by clausification #[4]): ∀ (a_1 : a → a → Prop) (a_2 : a),
% 3.72/3.91 Eq (∀ (Xy : a), Iff (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) Xy)) False
% 3.72/3.91 Clause #6 (by clausification #[5]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a),
% 3.72/3.91 Eq
% 3.72/3.91 (Not
% 3.72/3.91 (Iff (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 2 a_1 a_2 a_3))))
% 3.72/3.91 True
% 3.72/3.91 Clause #7 (by clausification #[6]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a),
% 3.72/3.91 Eq (Iff (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 2 a_1 a_2 a_3)))
% 3.72/3.91 False
% 3.72/3.91 Clause #8 (by clausification #[7]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a),
% 3.72/3.91 Or (Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3)) False)
% 3.72/3.91 (Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3)) False)
% 3.72/3.91 Clause #9 (by clausification #[7]): ∀ (a_1 : a → a → Prop) (a_2 a_3 : a),
% 3.72/3.91 Or (Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3)) True)
% 3.72/3.91 (Eq (skS.0 0 a_1 (skS.0 1 a_1 a_2) (skS.0 2 a_1 a_2 a_3)) True)
% 3.72/3.91 Clause #10 (by eliminate duplicate literals #[8]): ∀ (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)) False
% 3.72/3.91 Clause #11 (by eliminate duplicate literals #[9]): ∀ (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
% 3.72/3.91 Clause #12 (by superposition #[11, 10]): Eq True False
% 3.72/3.91 Clause #13 (by clausification #[12]): False
% 3.72/3.91 SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------