TSTP Solution File: SET143^5 by Duper---1.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SET143^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n015.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 14:45:49 EDT 2023
% Result : Theorem 3.58s 3.87s
% Output : Proof 3.58s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : SET143^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.15 % Command : duper %s
% 0.14/0.36 % Computer : n015.cluster.edu
% 0.14/0.36 % Model : x86_64 x86_64
% 0.14/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.36 % Memory : 8042.1875MB
% 0.14/0.36 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.36 % CPULimit : 300
% 0.14/0.36 % WCLimit : 300
% 0.14/0.36 % DateTime : Sat Aug 26 10:47:22 EDT 2023
% 0.14/0.36 % CPUTime :
% 3.58/3.87 SZS status Theorem for theBenchmark.p
% 3.58/3.87 SZS output start Proof for theBenchmark.p
% 3.58/3.87 Clause #0 (by assumption #[]): Eq (Not (∀ (X Y Z : a → Prop), Eq (fun Xx => And (And (X Xx) (Y Xx)) (Z Xx)) fun Xx => And (And (X Xx) (Y Xx)) (Z Xx)))
% 3.58/3.87 True
% 3.58/3.87 Clause #1 (by clausification #[0]): Eq (∀ (X Y Z : a → Prop), Eq (fun Xx => And (And (X Xx) (Y Xx)) (Z Xx)) fun Xx => And (And (X Xx) (Y Xx)) (Z Xx)) False
% 3.58/3.87 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop),
% 3.58/3.87 Eq
% 3.58/3.87 (Not
% 3.58/3.87 (∀ (Y Z : a → Prop),
% 3.58/3.87 Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (Z Xx)) fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (Z Xx)))
% 3.58/3.87 True
% 3.58/3.87 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop),
% 3.58/3.87 Eq
% 3.58/3.87 (∀ (Y Z : a → Prop),
% 3.58/3.87 Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (Z Xx)) fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (Z Xx))
% 3.58/3.87 False
% 3.58/3.87 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 3.58/3.87 Eq
% 3.58/3.87 (Not
% 3.58/3.87 (∀ (Z : a → Prop),
% 3.58/3.87 Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Z Xx)) fun Xx =>
% 3.58/3.87 And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Z Xx)))
% 3.58/3.87 True
% 3.58/3.87 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 3.58/3.87 Eq
% 3.58/3.87 (∀ (Z : a → Prop),
% 3.58/3.87 Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Z Xx)) fun Xx =>
% 3.58/3.87 And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Z Xx))
% 3.58/3.87 False
% 3.58/3.87 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.58/3.87 Eq
% 3.58/3.87 (Not
% 3.58/3.87 (Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (skS.0 2 a_1 a_2 a_3 Xx)) fun Xx =>
% 3.58/3.87 And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (skS.0 2 a_1 a_2 a_3 Xx)))
% 3.58/3.87 True
% 3.58/3.87 Clause #7 (by clausification #[6]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.58/3.87 Eq
% 3.58/3.87 (Eq (fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (skS.0 2 a_1 a_2 a_3 Xx)) fun Xx =>
% 3.58/3.87 And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (skS.0 2 a_1 a_2 a_3 Xx))
% 3.58/3.87 False
% 3.58/3.87 Clause #8 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.58/3.87 Ne (fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (skS.0 2 a_1 a_2 a_3 Xx)) fun Xx =>
% 3.58/3.87 And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (skS.0 2 a_1 a_2 a_3 Xx)
% 3.58/3.87 Clause #9 (by eliminate resolved literals #[8]): False
% 3.58/3.87 SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------