%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SYN969+1 : TPTP v9.2.0. Released v3.1.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n020.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 08:08:20 PM UTC 2025 % Result : Theorem 5.43s 5.65s % Output : Proof 5.43s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.04/0.08 % Problem : SYN969+1 : TPTP v9.2.0. Released v3.1.0. % 0.04/0.09 % Command : duper %s % 0.09/0.29 % Computer : n020.cluster.edu % 0.09/0.29 % Model : x86_64 x86_64 % 0.09/0.29 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.09/0.29 % Memory : 8042.1875MB % 0.09/0.29 % OS : Linux 3.10.0-693.el7.x86_64 % 0.09/0.29 % CPULimit : 300 % 0.09/0.29 % WCLimit : 300 % 0.09/0.29 % DateTime : Thu Oct 2 15:40:08 EDT 2025 % 0.09/0.30 % CPUTime : % 5.43/5.65 SZS status Theorem for theBenchmark.p % 5.43/5.65 SZS output start Proof for theBenchmark.p % 5.43/5.65 Clause #0 (by assumption #[]): Eq (Not (∀ (B : Iota), And (∀ (X : Iota), p X → q X) (r B) → (∀ (Y : Iota), r Y → p Y) → q B)) True % 5.43/5.65 Clause #1 (by clausification #[0]): Eq (∀ (B : Iota), And (∀ (X : Iota), p X → q X) (r B) → (∀ (Y : Iota), r Y → p Y) → q B) False % 5.43/5.65 Clause #2 (by clausification #[1]): ∀ (a : Iota), Eq (Not (And (∀ (X : Iota), p X → q X) (r (skS.0 0 a)) → (∀ (Y : Iota), r Y → p Y) → q (skS.0 0 a))) True % 5.43/5.65 Clause #3 (by clausification #[2]): ∀ (a : Iota), Eq (And (∀ (X : Iota), p X → q X) (r (skS.0 0 a)) → (∀ (Y : Iota), r Y → p Y) → q (skS.0 0 a)) False % 5.43/5.65 Clause #4 (by clausification #[3]): ∀ (a : Iota), Eq (And (∀ (X : Iota), p X → q X) (r (skS.0 0 a))) True % 5.43/5.65 Clause #5 (by clausification #[3]): ∀ (a : Iota), Eq ((∀ (Y : Iota), r Y → p Y) → q (skS.0 0 a)) False % 5.43/5.65 Clause #6 (by clausification #[4]): ∀ (a : Iota), Eq (r (skS.0 0 a)) True % 5.43/5.65 Clause #7 (by clausification #[4]): Eq (∀ (X : Iota), p X → q X) True % 5.43/5.65 Clause #8 (by clausification #[7]): ∀ (a : Iota), Eq (p a → q a) True % 5.43/5.65 Clause #9 (by clausification #[8]): ∀ (a : Iota), Or (Eq (p a) False) (Eq (q a) True) % 5.43/5.65 Clause #10 (by clausification #[5]): Eq (∀ (Y : Iota), r Y → p Y) True % 5.43/5.65 Clause #11 (by clausification #[5]): ∀ (a : Iota), Eq (q (skS.0 0 a)) False % 5.43/5.65 Clause #12 (by clausification #[10]): ∀ (a : Iota), Eq (r a → p a) True % 5.43/5.65 Clause #13 (by clausification #[12]): ∀ (a : Iota), Or (Eq (r a) False) (Eq (p a) True) % 5.43/5.65 Clause #14 (by superposition #[13, 6]): ∀ (a : Iota), Or (Eq (p (skS.0 0 a)) True) (Eq False True) % 5.43/5.65 Clause #15 (by clausification #[14]): ∀ (a : Iota), Eq (p (skS.0 0 a)) True % 5.43/5.65 Clause #16 (by superposition #[15, 9]): ∀ (a : Iota), Or (Eq True False) (Eq (q (skS.0 0 a)) True) % 5.43/5.65 Clause #17 (by clausification #[16]): ∀ (a : Iota), Eq (q (skS.0 0 a)) True % 5.43/5.65 Clause #18 (by superposition #[17, 11]): Eq True False % 5.43/5.65 Clause #19 (by clausification #[18]): False % 5.43/5.65 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------