TSTP Solution File: SYO212^5 by Duper---1.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SYO212^5 : 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 : Fri Sep 1 04:21:52 EDT 2023
% Result : Theorem 3.35s 3.63s
% Output : Proof 3.35s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SYO212^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.13 % Command : duper %s
% 0.14/0.35 % Computer : n015.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 02:08:52 EDT 2023
% 0.14/0.35 % CPUTime :
% 3.35/3.63 SZS status Theorem for theBenchmark.p
% 3.35/3.63 SZS output start Proof for theBenchmark.p
% 3.35/3.63 Clause #0 (by assumption #[]): Eq (Not (∀ (Xp : Prop → Prop), And (Xp a) (Xp b) → Xp (And a b))) True
% 3.35/3.63 Clause #1 (by clausification #[0]): Eq (∀ (Xp : Prop → Prop), And (Xp a) (Xp b) → Xp (And a b)) False
% 3.35/3.63 Clause #2 (by clausification #[1]): ∀ (a_1 : Prop → Prop), Eq (Not (And (skS.0 0 a_1 a) (skS.0 0 a_1 b) → skS.0 0 a_1 (And a b))) True
% 3.35/3.63 Clause #3 (by clausification #[2]): ∀ (a_1 : Prop → Prop), Eq (And (skS.0 0 a_1 a) (skS.0 0 a_1 b) → skS.0 0 a_1 (And a b)) False
% 3.35/3.63 Clause #4 (by clausification #[3]): ∀ (a_1 : Prop → Prop), Eq (And (skS.0 0 a_1 a) (skS.0 0 a_1 b)) True
% 3.35/3.63 Clause #5 (by clausification #[3]): ∀ (a_1 : Prop → Prop), Eq (skS.0 0 a_1 (And a b)) False
% 3.35/3.63 Clause #6 (by clausification #[4]): ∀ (a : Prop → Prop), Eq (skS.0 0 a b) True
% 3.35/3.63 Clause #7 (by clausification #[4]): ∀ (a_1 : Prop → Prop), Eq (skS.0 0 a_1 a) True
% 3.35/3.63 Clause #8 (by identity loobHoist #[6]): ∀ (a : Prop → Prop), Or (Eq (skS.0 0 a True) True) (Eq b False)
% 3.35/3.63 Clause #9 (by identity boolHoist #[6]): ∀ (a : Prop → Prop), Or (Eq (skS.0 0 a False) True) (Eq b True)
% 3.35/3.63 Clause #11 (by identity boolHoist #[7]): ∀ (a_1 : Prop → Prop), Or (Eq (skS.0 0 a_1 False) True) (Eq a True)
% 3.35/3.63 Clause #12 (by identity loobHoist #[5]): ∀ (a_1 : Prop → Prop), Or (Eq (skS.0 0 a_1 True) False) (Eq (And a b) False)
% 3.35/3.63 Clause #13 (by identity boolHoist #[5]): ∀ (a_1 : Prop → Prop), Or (Eq (skS.0 0 a_1 False) False) (Eq (And a b) True)
% 3.35/3.63 Clause #14 (by clausification #[12]): ∀ (a_1 : Prop → Prop), Or (Eq (skS.0 0 a_1 True) False) (Or (Eq a False) (Eq b False))
% 3.35/3.63 Clause #15 (by clausification #[13]): ∀ (a : Prop → Prop), Or (Eq (skS.0 0 a False) False) (Eq b True)
% 3.35/3.63 Clause #16 (by clausification #[13]): ∀ (a_1 : Prop → Prop), Or (Eq (skS.0 0 a_1 False) False) (Eq a True)
% 3.35/3.63 Clause #17 (by superposition #[15, 9]): Or (Eq b True) (Or (Eq False True) (Eq b True))
% 3.35/3.63 Clause #21 (by clausification #[17]): Or (Eq b True) (Eq b True)
% 3.35/3.63 Clause #22 (by eliminate duplicate literals #[21]): Eq b True
% 3.35/3.63 Clause #23 (by backward demodulation #[22, 8]): ∀ (a : Prop → Prop), Or (Eq (skS.0 0 a True) True) (Eq True False)
% 3.35/3.63 Clause #25 (by backward demodulation #[22, 14]): ∀ (a_1 : Prop → Prop), Or (Eq (skS.0 0 a_1 True) False) (Or (Eq a False) (Eq True False))
% 3.35/3.63 Clause #27 (by clausification #[23]): ∀ (a : Prop → Prop), Eq (skS.0 0 a True) True
% 3.35/3.63 Clause #28 (by superposition #[16, 11]): Or (Eq a True) (Or (Eq False True) (Eq a True))
% 3.35/3.63 Clause #30 (by clausification #[28]): Or (Eq a True) (Eq a True)
% 3.35/3.63 Clause #31 (by eliminate duplicate literals #[30]): Eq a True
% 3.35/3.63 Clause #32 (by clausification #[25]): ∀ (a_1 : Prop → Prop), Or (Eq (skS.0 0 a_1 True) False) (Eq a False)
% 3.35/3.63 Clause #33 (by forward demodulation #[32, 31]): ∀ (a : Prop → Prop), Or (Eq (skS.0 0 a True) False) (Eq True False)
% 3.35/3.63 Clause #34 (by clausification #[33]): ∀ (a : Prop → Prop), Eq (skS.0 0 a True) False
% 3.35/3.63 Clause #35 (by superposition #[34, 27]): Eq False True
% 3.35/3.63 Clause #36 (by clausification #[35]): False
% 3.35/3.63 SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------