TSTP Solution File: SET585^5 by Duper---1.0
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% File : Duper---1.0
% Problem : SET585^5 : 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 : Thu Aug 31 14:46:54 EDT 2023
% Result : Theorem 3.86s 4.02s
% Output : Proof 3.86s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : SET585^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14 % Command : duper %s
% 0.14/0.34 % Computer : n015.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Sat Aug 26 15:03:38 EDT 2023
% 0.14/0.35 % CPUTime :
% 3.86/4.02 SZS status Theorem for theBenchmark.p
% 3.86/4.02 SZS output start Proof for theBenchmark.p
% 3.86/4.02 Clause #0 (by assumption #[]): Eq (Not (∀ (X Y Z : a → Prop) (Xx : a), And (X Xx) (Y Xx) → Or (X Xx) (Z Xx))) True
% 3.86/4.02 Clause #1 (by clausification #[0]): Eq (∀ (X Y Z : a → Prop) (Xx : a), And (X Xx) (Y Xx) → Or (X Xx) (Z Xx)) False
% 3.86/4.02 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop),
% 3.86/4.02 Eq (Not (∀ (Y Z : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (Y Xx) → Or (skS.0 0 a_1 Xx) (Z Xx))) True
% 3.86/4.02 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop), Eq (∀ (Y Z : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (Y Xx) → Or (skS.0 0 a_1 Xx) (Z Xx)) False
% 3.86/4.02 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 3.86/4.02 Eq (Not (∀ (Z : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → Or (skS.0 0 a_1 Xx) (Z Xx))) True
% 3.86/4.02 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 3.86/4.02 Eq (∀ (Z : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → Or (skS.0 0 a_1 Xx) (Z Xx)) False
% 3.86/4.02 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.86/4.02 Eq (Not (∀ (Xx : a), And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → Or (skS.0 0 a_1 Xx) (skS.0 2 a_1 a_2 a_3 Xx))) True
% 3.86/4.02 Clause #7 (by clausification #[6]): ∀ (a_1 a_2 a_3 : a → Prop),
% 3.86/4.02 Eq (∀ (Xx : a), And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → Or (skS.0 0 a_1 Xx) (skS.0 2 a_1 a_2 a_3 Xx)) False
% 3.86/4.02 Clause #8 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a),
% 3.86/4.02 Eq
% 3.86/4.02 (Not
% 3.86/4.02 (And (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3 a_4)) →
% 3.86/4.02 Or (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4))))
% 3.86/4.02 True
% 3.86/4.02 Clause #9 (by clausification #[8]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a),
% 3.86/4.02 Eq
% 3.86/4.02 (And (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3 a_4)) →
% 3.86/4.02 Or (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4)))
% 3.86/4.02 False
% 3.86/4.02 Clause #10 (by clausification #[9]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a),
% 3.86/4.02 Eq (And (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3 a_4))) True
% 3.86/4.02 Clause #11 (by clausification #[9]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a),
% 3.86/4.02 Eq (Or (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4))) False
% 3.86/4.02 Clause #13 (by clausification #[10]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) True
% 3.86/4.02 Clause #15 (by clausification #[11]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) False
% 3.86/4.02 Clause #16 (by superposition #[15, 13]): Eq False True
% 3.86/4.02 Clause #17 (by clausification #[16]): False
% 3.86/4.02 SZS output end Proof for theBenchmark.p
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