TSTP Solution File: SYO348^5 by Duper---1.0
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% File : Duper---1.0
% Problem : SYO348^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n016.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 04:22:13 EDT 2023
% Result : Theorem 3.63s 3.80s
% Output : Proof 3.63s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SYO348^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.13 % Command : duper %s
% 0.14/0.35 % Computer : n016.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 05:26:56 EDT 2023
% 0.14/0.35 % CPUTime :
% 3.63/3.80 SZS status Theorem for theBenchmark.p
% 3.63/3.80 SZS output start Proof for theBenchmark.p
% 3.63/3.80 Clause #0 (by assumption #[]): Eq (Not (Iff a b → ∀ (P : Prop → Prop), P a → P b)) True
% 3.63/3.80 Clause #1 (by clausification #[0]): Eq (Iff a b → ∀ (P : Prop → Prop), P a → P b) False
% 3.63/3.80 Clause #2 (by clausification #[1]): Eq (Iff a b) True
% 3.63/3.80 Clause #3 (by clausification #[1]): Eq (∀ (P : Prop → Prop), P a → P b) False
% 3.63/3.80 Clause #4 (by clausification #[2]): Or (Eq a True) (Eq b False)
% 3.63/3.80 Clause #5 (by clausification #[2]): Or (Eq a False) (Eq b True)
% 3.63/3.80 Clause #6 (by clausification #[3]): ∀ (a_1 : Prop → Prop), Eq (Not (skS.0 0 a_1 a → skS.0 0 a_1 b)) True
% 3.63/3.80 Clause #7 (by clausification #[6]): ∀ (a_1 : Prop → Prop), Eq (skS.0 0 a_1 a → skS.0 0 a_1 b) False
% 3.63/3.80 Clause #8 (by clausification #[7]): ∀ (a_1 : Prop → Prop), Eq (skS.0 0 a_1 a) True
% 3.63/3.80 Clause #9 (by clausification #[7]): ∀ (a : Prop → Prop), Eq (skS.0 0 a b) False
% 3.63/3.80 Clause #10 (by identity loobHoist #[8]): ∀ (a_1 : Prop → Prop), Or (Eq (skS.0 0 a_1 True) True) (Eq a False)
% 3.63/3.80 Clause #11 (by identity boolHoist #[8]): ∀ (a_1 : Prop → Prop), Or (Eq (skS.0 0 a_1 False) True) (Eq a True)
% 3.63/3.80 Clause #12 (by identity loobHoist #[9]): ∀ (a : Prop → Prop), Or (Eq (skS.0 0 a True) False) (Eq b False)
% 3.63/3.80 Clause #13 (by identity boolHoist #[9]): ∀ (a : Prop → Prop), Or (Eq (skS.0 0 a False) False) (Eq b True)
% 3.63/3.80 Clause #14 (by superposition #[13, 11]): Or (Eq b True) (Or (Eq False True) (Eq a True))
% 3.63/3.80 Clause #15 (by clausification #[14]): Or (Eq b True) (Eq a True)
% 3.63/3.80 Clause #16 (by superposition #[15, 4]): Or (Eq a True) (Or (Eq a True) (Eq True False))
% 3.63/3.80 Clause #17 (by clausification #[16]): Or (Eq a True) (Eq a True)
% 3.63/3.80 Clause #18 (by eliminate duplicate literals #[17]): Eq a True
% 3.63/3.80 Clause #20 (by backward demodulation #[18, 5]): Or (Eq True False) (Eq b True)
% 3.63/3.80 Clause #21 (by backward demodulation #[18, 10]): ∀ (a : Prop → Prop), Or (Eq (skS.0 0 a True) True) (Eq True False)
% 3.63/3.80 Clause #24 (by clausification #[20]): Eq b True
% 3.63/3.80 Clause #25 (by backward demodulation #[24, 12]): ∀ (a : Prop → Prop), Or (Eq (skS.0 0 a True) False) (Eq True False)
% 3.63/3.80 Clause #27 (by clausification #[21]): ∀ (a : Prop → Prop), Eq (skS.0 0 a True) True
% 3.63/3.80 Clause #28 (by clausification #[25]): ∀ (a : Prop → Prop), Eq (skS.0 0 a True) False
% 3.63/3.80 Clause #29 (by superposition #[28, 27]): Eq False True
% 3.63/3.80 Clause #30 (by clausification #[29]): False
% 3.63/3.80 SZS output end Proof for theBenchmark.p
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