TSTP Solution File: SET185^5 by Duper---1.0
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% File : Duper---1.0
% Problem : SET185^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n002.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 14:45:58 EDT 2023
% Result : Theorem 3.49s 3.73s
% Output : Proof 3.49s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12 % Problem : SET185^5 : TPTP v8.1.2. Released v4.0.0.
% 0.06/0.13 % Command : duper %s
% 0.13/0.34 % Computer : n002.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Sat Aug 26 11:45:48 EDT 2023
% 0.19/0.34 % CPUTime :
% 3.49/3.73 SZS status Theorem for theBenchmark.p
% 3.49/3.73 SZS output start Proof for theBenchmark.p
% 3.49/3.73 Clause #0 (by assumption #[]): Eq (Not (∀ (X Y : a → Prop), (∀ (Xx : a), X Xx → Y Xx) → ∀ (Xx : a), Or (X Xx) (Y Xx) → Y Xx)) True
% 3.49/3.73 Clause #1 (by clausification #[0]): Eq (∀ (X Y : a → Prop), (∀ (Xx : a), X Xx → Y Xx) → ∀ (Xx : a), Or (X Xx) (Y Xx) → Y Xx) False
% 3.49/3.73 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop),
% 3.49/3.73 Eq (Not (∀ (Y : a → Prop), (∀ (Xx : a), skS.0 0 a_1 Xx → Y Xx) → ∀ (Xx : a), Or (skS.0 0 a_1 Xx) (Y Xx) → Y Xx)) True
% 3.49/3.73 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop),
% 3.49/3.73 Eq (∀ (Y : a → Prop), (∀ (Xx : a), skS.0 0 a_1 Xx → Y Xx) → ∀ (Xx : a), Or (skS.0 0 a_1 Xx) (Y Xx) → Y Xx) False
% 3.49/3.73 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 3.49/3.73 Eq
% 3.49/3.73 (Not
% 3.49/3.73 ((∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 1 a_1 a_2 Xx) →
% 3.49/3.73 ∀ (Xx : a), Or (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → skS.0 1 a_1 a_2 Xx))
% 3.49/3.73 True
% 3.49/3.73 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 3.49/3.73 Eq
% 3.49/3.73 ((∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 1 a_1 a_2 Xx) →
% 3.49/3.73 ∀ (Xx : a), Or (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → skS.0 1 a_1 a_2 Xx)
% 3.49/3.73 False
% 3.49/3.73 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 : a → Prop), Eq (∀ (Xx : a), skS.0 0 a_1 Xx → skS.0 1 a_1 a_2 Xx) True
% 3.49/3.73 Clause #7 (by clausification #[5]): ∀ (a_1 a_2 : a → Prop), Eq (∀ (Xx : a), Or (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → skS.0 1 a_1 a_2 Xx) False
% 3.49/3.73 Clause #8 (by clausification #[6]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 : a → Prop), Eq (skS.0 0 a_1 a_2 → skS.0 1 a_1 a_3 a_2) True
% 3.49/3.73 Clause #9 (by clausification #[8]): ∀ (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) True)
% 3.49/3.73 Clause #10 (by clausification #[7]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.49/3.73 Eq
% 3.49/3.73 (Not
% 3.49/3.73 (Or (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)) →
% 3.49/3.73 skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)))
% 3.49/3.73 True
% 3.49/3.73 Clause #11 (by clausification #[10]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.49/3.73 Eq
% 3.49/3.73 (Or (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)) →
% 3.49/3.73 skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))
% 3.49/3.73 False
% 3.49/3.73 Clause #12 (by clausification #[11]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.49/3.73 Eq (Or (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
% 3.49/3.73 Clause #13 (by clausification #[11]): ∀ (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)) False
% 3.49/3.73 Clause #14 (by clausification #[12]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.49/3.73 Or (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) True) (Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)) True)
% 3.49/3.73 Clause #15 (by superposition #[13, 14]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.49/3.73 Or (Eq (skS.0 0 (fun x => a_1 x) (skS.0 2 (fun x => a_1 x) (fun x => a_2 x) a_3)) True) (Eq True False)
% 3.49/3.73 Clause #16 (by betaEtaReduce #[15]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Or (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) True) (Eq True False)
% 3.49/3.73 Clause #17 (by clausification #[16]): ∀ (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
% 3.49/3.73 Clause #19 (by superposition #[17, 9]): ∀ (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)) True)
% 3.49/3.73 Clause #20 (by clausification #[19]): ∀ (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)) True
% 3.49/3.73 Clause #21 (by superposition #[20, 13]): Eq True False
% 3.49/3.73 Clause #22 (by clausification #[21]): False
% 3.49/3.73 SZS output end Proof for theBenchmark.p
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