%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SET631^5 : TPTP v9.2.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n017.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 Oct 3 08:00:18 PM UTC 2025 % Result : Theorem 3.97s 4.15s % Output : Proof 3.97s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.10/0.13 % Problem : SET631^5 : TPTP v9.2.0. Released v4.0.0. % 0.10/0.14 % Command : duper %s % 0.14/0.36 % Computer : n017.cluster.edu % 0.14/0.36 % Model : x86_64 x86_64 % 0.14/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.14/0.36 % Memory : 8042.1875MB % 0.14/0.36 % OS : Linux 3.10.0-693.el7.x86_64 % 0.14/0.36 % CPULimit : 300 % 0.14/0.36 % WCLimit : 300 % 0.14/0.36 % DateTime : Thu Oct 2 14:09:23 EDT 2025 % 0.14/0.36 % CPUTime : % 3.97/4.15 SZS status Theorem for theBenchmark.p % 3.97/4.15 SZS output start Proof for theBenchmark.p % 3.97/4.15 Clause #0 (by assumption #[]): Eq % 3.97/4.15 (Not % 3.97/4.15 (∀ (X Y Z : a → Prop), % 3.97/4.15 (Exists fun Xx => And (And (X Xx) (Y Xx)) (Not (Z Xx))) → Exists fun Xx => And (X Xx) (Y Xx))) % 3.97/4.15 True % 3.97/4.15 Clause #1 (by clausification #[0]): Eq (∀ (X Y Z : a → Prop), (Exists fun Xx => And (And (X Xx) (Y Xx)) (Not (Z Xx))) → Exists fun Xx => And (X Xx) (Y Xx)) % 3.97/4.15 False % 3.97/4.15 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop), % 3.97/4.15 Eq % 3.97/4.15 (Not % 3.97/4.15 (∀ (Y Z : a → Prop), % 3.97/4.15 (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (Not (Z Xx))) → % 3.97/4.15 Exists fun Xx => And (skS.0 0 a_1 Xx) (Y Xx))) % 3.97/4.15 True % 3.97/4.15 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop), % 3.97/4.15 Eq % 3.97/4.15 (∀ (Y Z : a → Prop), % 3.97/4.15 (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (Not (Z Xx))) → Exists fun Xx => And (skS.0 0 a_1 Xx) (Y Xx)) % 3.97/4.15 False % 3.97/4.15 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop), % 3.97/4.15 Eq % 3.97/4.15 (Not % 3.97/4.15 (∀ (Z : a → Prop), % 3.97/4.15 (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (Z Xx))) → % 3.97/4.15 Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx))) % 3.97/4.15 True % 3.97/4.15 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop), % 3.97/4.15 Eq % 3.97/4.15 (∀ (Z : a → Prop), % 3.97/4.15 (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (Z Xx))) → % 3.97/4.15 Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) % 3.97/4.15 False % 3.97/4.15 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 a_3 : a → Prop), % 3.97/4.15 Eq % 3.97/4.15 (Not % 3.97/4.15 ((Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))) → % 3.97/4.15 Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx))) % 3.97/4.15 True % 3.97/4.15 Clause #7 (by clausification #[6]): ∀ (a_1 a_2 a_3 : a → Prop), % 3.97/4.15 Eq % 3.97/4.15 ((Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))) → % 3.97/4.15 Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) % 3.97/4.15 False % 3.97/4.15 Clause #8 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop), % 3.97/4.15 Eq (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (Not (skS.0 2 a_1 a_2 a_3 Xx))) True % 3.97/4.15 Clause #9 (by clausification #[7]): ∀ (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)) False % 3.97/4.15 Clause #10 (by clausification #[8]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), % 3.97/4.15 Eq % 3.97/4.15 (And (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))) % 3.97/4.15 (Not (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4)))) % 3.97/4.15 True % 3.97/4.15 Clause #12 (by clausification #[10]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), % 3.97/4.15 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 % 3.97/4.15 Clause #14 (by clausification #[9]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 : a → Prop), Eq (And (skS.0 0 a_1 a_2) (skS.0 1 a_1 a_3 a_2)) False % 3.97/4.15 Clause #15 (by clausification #[14]): ∀ (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) False) % 3.97/4.15 Clause #16 (by clausification #[12]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3 a_4)) True % 3.97/4.15 Clause #17 (by clausification #[12]): ∀ (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 % 3.97/4.15 Clause #18 (by superposition #[17, 15]): ∀ (a_1 a_2 a_3 a_4 : a → Prop) (a_5 : a), Or (Eq True False) (Eq (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_3 a_4 a_5)) False) % 3.97/4.15 Clause #19 (by clausification #[18]): ∀ (a_1 a_2 a_3 a_4 : a → Prop) (a_5 : a), Eq (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_3 a_4 a_5)) False % 3.97/4.15 Clause #20 (by superposition #[19, 16]): Eq False True % 3.97/4.15 Clause #21 (by clausification #[20]): False % 3.97/4.15 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------