TSTP Solution File: SET625^5 by Duper---1.0
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%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SET625^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n010.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 14:47:08 EDT 2023
% Result : Theorem 3.64s 3.81s
% Output : Proof 3.64s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.11 % Problem : SET625^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.13 % Command : duper %s
% 0.12/0.33 % Computer : n010.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Sat Aug 26 09:37:50 EDT 2023
% 0.12/0.33 % CPUTime :
% 3.64/3.81 SZS status Theorem for theBenchmark.p
% 3.64/3.81 SZS output start Proof for theBenchmark.p
% 3.64/3.81 Clause #0 (by assumption #[]): Eq
% 3.64/3.81 (Not
% 3.64/3.81 (∀ (X Y Z : a → Prop),
% 3.64/3.81 And (Exists fun Xx => And (X Xx) (Y Xx)) (∀ (Xx : a), Y Xx → Z Xx) → Exists fun Xx => And (X Xx) (Z Xx)))
% 3.64/3.81 True
% 3.64/3.81 Clause #1 (by clausification #[0]): Eq
% 3.64/3.81 (∀ (X Y Z : a → Prop),
% 3.64/3.81 And (Exists fun Xx => And (X Xx) (Y Xx)) (∀ (Xx : a), Y Xx → Z Xx) → Exists fun Xx => And (X Xx) (Z Xx))
% 3.64/3.81 False
% 3.64/3.81 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop),
% 3.64/3.81 Eq
% 3.64/3.81 (Not
% 3.64/3.81 (∀ (Y Z : a → Prop),
% 3.64/3.81 And (Exists fun Xx => And (skS.0 0 a_1 Xx) (Y Xx)) (∀ (Xx : a), Y Xx → Z Xx) →
% 3.64/3.81 Exists fun Xx => And (skS.0 0 a_1 Xx) (Z Xx)))
% 3.64/3.81 True
% 3.64/3.81 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop),
% 3.64/3.81 Eq
% 3.64/3.81 (∀ (Y Z : a → Prop),
% 3.64/3.81 And (Exists fun Xx => And (skS.0 0 a_1 Xx) (Y Xx)) (∀ (Xx : a), Y Xx → Z Xx) →
% 3.64/3.81 Exists fun Xx => And (skS.0 0 a_1 Xx) (Z Xx))
% 3.64/3.81 False
% 3.64/3.81 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 3.64/3.81 Eq
% 3.64/3.81 (Not
% 3.64/3.81 (∀ (Z : a → Prop),
% 3.64/3.81 And (Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → Z Xx) →
% 3.64/3.81 Exists fun Xx => And (skS.0 0 a_1 Xx) (Z Xx)))
% 3.64/3.81 True
% 3.64/3.81 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 3.64/3.81 Eq
% 3.64/3.81 (∀ (Z : a → Prop),
% 3.64/3.81 And (Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → Z Xx) →
% 3.64/3.81 Exists fun Xx => And (skS.0 0 a_1 Xx) (Z Xx))
% 3.64/3.81 False
% 3.64/3.81 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.64/3.81 Eq
% 3.64/3.81 (Not
% 3.64/3.81 (And (Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx))
% 3.64/3.81 (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → skS.0 2 a_1 a_2 a_3 Xx) →
% 3.64/3.81 Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 2 a_1 a_2 a_3 Xx)))
% 3.64/3.81 True
% 3.64/3.81 Clause #7 (by clausification #[6]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.64/3.81 Eq
% 3.64/3.81 (And (Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx))
% 3.64/3.81 (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → skS.0 2 a_1 a_2 a_3 Xx) →
% 3.64/3.81 Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 2 a_1 a_2 a_3 Xx))
% 3.64/3.81 False
% 3.64/3.81 Clause #8 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.64/3.81 Eq
% 3.64/3.81 (And (Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx))
% 3.64/3.81 (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → skS.0 2 a_1 a_2 a_3 Xx))
% 3.64/3.81 True
% 3.64/3.81 Clause #9 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop), Eq (Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 2 a_1 a_2 a_3 Xx)) False
% 3.64/3.81 Clause #10 (by clausification #[8]): ∀ (a_1 a_2 a_3 : a → Prop), Eq (∀ (Xx : a), skS.0 1 a_1 a_2 Xx → skS.0 2 a_1 a_2 a_3 Xx) True
% 3.64/3.81 Clause #11 (by clausification #[8]): ∀ (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)) True
% 3.64/3.81 Clause #12 (by clausification #[10]): ∀ (a_1 a_2 : a → Prop) (a_3 : a) (a_4 : a → Prop), Eq (skS.0 1 a_1 a_2 a_3 → skS.0 2 a_1 a_2 a_4 a_3) True
% 3.64/3.81 Clause #13 (by clausification #[12]): ∀ (a_1 a_2 : a → Prop) (a_3 : a) (a_4 : a → Prop),
% 3.64/3.81 Or (Eq (skS.0 1 a_1 a_2 a_3) False) (Eq (skS.0 2 a_1 a_2 a_4 a_3) True)
% 3.64/3.81 Clause #14 (by clausification #[11]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.64/3.81 Eq (And (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3))) True
% 3.64/3.81 Clause #15 (by clausification #[14]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3)) True
% 3.64/3.81 Clause #16 (by clausification #[14]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3)) True
% 3.64/3.81 Clause #17 (by superposition #[15, 13]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Or (Eq True False) (Eq (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_4)) True)
% 3.64/3.81 Clause #18 (by clausification #[17]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_4)) True
% 3.64/3.81 Clause #19 (by clausification #[9]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 a_4 : a → Prop), Eq (And (skS.0 0 a_1 a_2) (skS.0 2 a_1 a_3 a_4 a_2)) False
% 3.64/3.81 Clause #20 (by clausification #[19]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 a_4 : a → Prop), Or (Eq (skS.0 0 a_1 a_2) False) (Eq (skS.0 2 a_1 a_3 a_4 a_2) False)
% 3.64/3.82 Clause #21 (by superposition #[20, 16]): ∀ (a_1 a_2 a_3 a_4 : a → Prop) (a_5 : a),
% 3.64/3.82 Or (Eq (skS.0 2 (fun x => a_1 x) a_2 a_3 (skS.0 3 a_1 a_4 a_5)) False) (Eq False True)
% 3.64/3.82 Clause #22 (by betaEtaReduce #[21]): ∀ (a_1 a_2 a_3 a_4 : a → Prop) (a_5 : a), Or (Eq (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_4 a_5)) False) (Eq False True)
% 3.64/3.82 Clause #23 (by clausification #[22]): ∀ (a_1 a_2 a_3 a_4 : a → Prop) (a_5 : a), Eq (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_4 a_5)) False
% 3.64/3.82 Clause #24 (by superposition #[23, 18]): Eq False True
% 3.64/3.82 Clause #25 (by clausification #[24]): False
% 3.64/3.82 SZS output end Proof for theBenchmark.p
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