TSTP Solution File: SET626^5 by Duper---1.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SET626^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n009.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 14:47:09 EDT 2023
% Result : Theorem 3.54s 3.71s
% Output : Proof 3.54s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.10/0.12 % Problem : SET626^5 : TPTP v8.1.2. Released v4.0.0.
% 0.10/0.13 % Command : duper %s
% 0.14/0.35 % Computer : n009.cluster.edu
% 0.14/0.35 % Model : x86_64 x86_64
% 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35 % Memory : 8042.1875MB
% 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35 % CPULimit : 300
% 0.14/0.35 % WCLimit : 300
% 0.14/0.35 % DateTime : Sat Aug 26 11:35:20 EDT 2023
% 0.14/0.35 % CPUTime :
% 3.54/3.71 SZS status Theorem for theBenchmark.p
% 3.54/3.71 SZS output start Proof for theBenchmark.p
% 3.54/3.71 Clause #0 (by assumption #[]): Eq (Not (∀ (X Y : a → Prop), (Exists fun Xx => And (And (X Xx) (Y Xx)) (cZ Xx)) → Exists fun Xx => And (X Xx) (Y Xx)))
% 3.54/3.71 True
% 3.54/3.71 Clause #1 (by clausification #[0]): Eq (∀ (X Y : a → Prop), (Exists fun Xx => And (And (X Xx) (Y Xx)) (cZ Xx)) → Exists fun Xx => And (X Xx) (Y Xx)) False
% 3.54/3.71 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop),
% 3.54/3.71 Eq
% 3.54/3.71 (Not
% 3.54/3.71 (∀ (Y : a → Prop),
% 3.54/3.71 (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (cZ Xx)) → Exists fun Xx => And (skS.0 0 a_1 Xx) (Y Xx)))
% 3.54/3.71 True
% 3.54/3.71 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop),
% 3.54/3.71 Eq
% 3.54/3.71 (∀ (Y : a → Prop),
% 3.54/3.71 (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (cZ Xx)) → Exists fun Xx => And (skS.0 0 a_1 Xx) (Y Xx))
% 3.54/3.71 False
% 3.54/3.71 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 3.54/3.71 Eq
% 3.54/3.71 (Not
% 3.54/3.71 ((Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (cZ Xx)) →
% 3.54/3.71 Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)))
% 3.54/3.71 True
% 3.54/3.71 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 3.54/3.71 Eq
% 3.54/3.71 ((Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (cZ Xx)) →
% 3.54/3.71 Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx))
% 3.54/3.71 False
% 3.54/3.71 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 : a → Prop), Eq (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (cZ Xx)) True
% 3.54/3.71 Clause #7 (by clausification #[5]): ∀ (a_1 a_2 : a → Prop), Eq (Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) False
% 3.54/3.71 Clause #8 (by clausification #[6]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.54/3.71 Eq (And (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) (cZ (skS.0 2 a_1 a_2 a_3)))
% 3.54/3.71 True
% 3.54/3.71 Clause #10 (by clausification #[8]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.54/3.71 Eq (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) True
% 3.54/3.71 Clause #11 (by clausification #[7]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 : a → Prop), Eq (And (skS.0 0 a_1 a_2) (skS.0 1 a_1 a_3 a_2)) False
% 3.54/3.71 Clause #12 (by clausification #[11]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 : a → Prop), Or (Eq (skS.0 0 a_1 a_2) False) (Eq (skS.0 1 a_1 a_3 a_2) False)
% 3.54/3.71 Clause #13 (by clausification #[10]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)) True
% 3.54/3.71 Clause #14 (by clausification #[10]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) True
% 3.54/3.71 Clause #15 (by superposition #[14, 12]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Or (Eq True False) (Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_3 a_4)) False)
% 3.54/3.71 Clause #16 (by clausification #[15]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_3 a_4)) False
% 3.54/3.71 Clause #17 (by superposition #[16, 13]): Eq False True
% 3.54/3.71 Clause #18 (by clausification #[17]): False
% 3.54/3.71 SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------