TSTP Solution File: SET159^5 by Duper---1.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SET159^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n004.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 14:45:52 EDT 2023
% Result : Theorem 4.47s 4.66s
% Output : Proof 4.47s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SET159^5 : TPTP v8.1.2. Released v4.0.0.
% 0.12/0.13 % Command : duper %s
% 0.12/0.33 % Computer : n004.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 : Sat Aug 26 09:57:23 EDT 2023
% 0.12/0.33 % CPUTime :
% 4.47/4.66 SZS status Theorem for theBenchmark.p
% 4.47/4.66 SZS output start Proof for theBenchmark.p
% 4.47/4.66 Clause #0 (by assumption #[]): Eq (Not (∀ (X Y Z : a → Prop), Eq (fun Xz => Or (Or (X Xz) (Y Xz)) (Z Xz)) fun Xz => Or (Or (X Xz) (Y Xz)) (Z Xz))) True
% 4.47/4.66 Clause #1 (by clausification #[0]): Eq (∀ (X Y Z : a → Prop), Eq (fun Xz => Or (Or (X Xz) (Y Xz)) (Z Xz)) fun Xz => Or (Or (X Xz) (Y Xz)) (Z Xz)) False
% 4.47/4.66 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop),
% 4.47/4.66 Eq
% 4.47/4.66 (Not
% 4.47/4.66 (∀ (Y Z : a → Prop),
% 4.47/4.66 Eq (fun Xz => Or (Or (skS.0 0 a_1 Xz) (Y Xz)) (Z Xz)) fun Xz => Or (Or (skS.0 0 a_1 Xz) (Y Xz)) (Z Xz)))
% 4.47/4.66 True
% 4.47/4.66 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop),
% 4.47/4.66 Eq
% 4.47/4.66 (∀ (Y Z : a → Prop),
% 4.47/4.66 Eq (fun Xz => Or (Or (skS.0 0 a_1 Xz) (Y Xz)) (Z Xz)) fun Xz => Or (Or (skS.0 0 a_1 Xz) (Y Xz)) (Z Xz))
% 4.47/4.66 False
% 4.47/4.66 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 4.47/4.66 Eq
% 4.47/4.66 (Not
% 4.47/4.66 (∀ (Z : a → Prop),
% 4.47/4.66 Eq (fun Xz => Or (Or (skS.0 0 a_1 Xz) (skS.0 1 a_1 a_2 Xz)) (Z Xz)) fun Xz =>
% 4.47/4.66 Or (Or (skS.0 0 a_1 Xz) (skS.0 1 a_1 a_2 Xz)) (Z Xz)))
% 4.47/4.66 True
% 4.47/4.66 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 4.47/4.66 Eq
% 4.47/4.66 (∀ (Z : a → Prop),
% 4.47/4.66 Eq (fun Xz => Or (Or (skS.0 0 a_1 Xz) (skS.0 1 a_1 a_2 Xz)) (Z Xz)) fun Xz =>
% 4.47/4.66 Or (Or (skS.0 0 a_1 Xz) (skS.0 1 a_1 a_2 Xz)) (Z Xz))
% 4.47/4.66 False
% 4.47/4.66 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 a_3 : a → Prop),
% 4.47/4.66 Eq
% 4.47/4.66 (Not
% 4.47/4.66 (Eq (fun Xz => Or (Or (skS.0 0 a_1 Xz) (skS.0 1 a_1 a_2 Xz)) (skS.0 2 a_1 a_2 a_3 Xz)) fun Xz =>
% 4.47/4.66 Or (Or (skS.0 0 a_1 Xz) (skS.0 1 a_1 a_2 Xz)) (skS.0 2 a_1 a_2 a_3 Xz)))
% 4.47/4.66 True
% 4.47/4.66 Clause #7 (by clausification #[6]): ∀ (a_1 a_2 a_3 : a → Prop),
% 4.47/4.66 Eq
% 4.47/4.66 (Eq (fun Xz => Or (Or (skS.0 0 a_1 Xz) (skS.0 1 a_1 a_2 Xz)) (skS.0 2 a_1 a_2 a_3 Xz)) fun Xz =>
% 4.47/4.66 Or (Or (skS.0 0 a_1 Xz) (skS.0 1 a_1 a_2 Xz)) (skS.0 2 a_1 a_2 a_3 Xz))
% 4.47/4.66 False
% 4.47/4.66 Clause #8 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop),
% 4.47/4.66 Ne (fun Xz => Or (Or (skS.0 0 a_1 Xz) (skS.0 1 a_1 a_2 Xz)) (skS.0 2 a_1 a_2 a_3 Xz)) fun Xz =>
% 4.47/4.66 Or (Or (skS.0 0 a_1 Xz) (skS.0 1 a_1 a_2 Xz)) (skS.0 2 a_1 a_2 a_3 Xz)
% 4.47/4.66 Clause #9 (by eliminate resolved literals #[8]): False
% 4.47/4.66 SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------