%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SYN983^1 : TPTP v9.2.0. Released v3.6.0. % 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 : Fri Oct 3 08:08:21 PM UTC 2025 % Result : Theorem 3.88s 4.07s % Output : Proof 3.88s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.07/0.13 % Problem : SYN983^1 : TPTP v9.2.0. Released v3.6.0. % 0.07/0.14 % Command : duper %s % 0.13/0.36 % Computer : n015.cluster.edu % 0.13/0.36 % Model : x86_64 x86_64 % 0.13/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.13/0.36 % Memory : 8042.1875MB % 0.13/0.36 % OS : Linux 3.10.0-693.el7.x86_64 % 0.13/0.36 % CPULimit : 300 % 0.13/0.36 % WCLimit : 300 % 0.13/0.36 % DateTime : Thu Oct 2 15:02:38 EDT 2025 % 0.13/0.36 % CPUTime : % 3.88/4.07 SZS status Theorem for theBenchmark.p % 3.88/4.07 SZS output start Proof for theBenchmark.p % 3.88/4.07 Clause #0 (by assumption #[]): Eq (Not (And (p a) (p b) → p (And a b))) True % 3.88/4.07 Clause #1 (by clausification #[0]): Eq (And (p a) (p b) → p (And a b)) False % 3.88/4.07 Clause #2 (by clausification #[1]): Eq (And (p a) (p b)) True % 3.88/4.07 Clause #3 (by clausification #[1]): Eq (p (And a b)) False % 3.88/4.07 Clause #4 (by clausification #[2]): Eq (p b) True % 3.88/4.07 Clause #5 (by clausification #[2]): Eq (p a) True % 3.88/4.07 Clause #7 (by identity boolHoist #[4]): Or (Eq (p False) True) (Eq b True) % 3.88/4.07 Clause #8 (by identity loobHoist #[5]): Or (Eq (p True) True) (Eq a False) % 3.88/4.07 Clause #9 (by identity boolHoist #[5]): Or (Eq (p False) True) (Eq a True) % 3.88/4.07 Clause #10 (by identity loobHoist #[3]): Or (Eq (p True) False) (Eq (And a b) False) % 3.88/4.07 Clause #11 (by identity boolHoist #[3]): Or (Eq (p False) False) (Eq (And a b) True) % 3.88/4.07 Clause #12 (by clausification #[10]): Or (Eq (p True) False) (Or (Eq a False) (Eq b False)) % 3.88/4.07 Clause #13 (by clausification #[11]): Or (Eq (p False) False) (Eq b True) % 3.88/4.07 Clause #14 (by clausification #[11]): Or (Eq (p False) False) (Eq a True) % 3.88/4.07 Clause #15 (by superposition #[13, 7]): Or (Eq b True) (Or (Eq False True) (Eq b True)) % 3.88/4.07 Clause #18 (by superposition #[14, 9]): Or (Eq a True) (Or (Eq False True) (Eq a True)) % 3.88/4.07 Clause #19 (by clausification #[18]): Or (Eq a True) (Eq a True) % 3.88/4.07 Clause #20 (by eliminate duplicate literals #[19]): Eq a True % 3.88/4.07 Clause #21 (by backward demodulation #[20, 8]): Or (Eq (p True) True) (Eq True False) % 3.88/4.07 Clause #22 (by backward demodulation #[20, 12]): Or (Eq (p True) False) (Or (Eq True False) (Eq b False)) % 3.88/4.07 Clause #25 (by clausification #[21]): Eq (p True) True % 3.88/4.07 Clause #26 (by clausification #[15]): Or (Eq b True) (Eq b True) % 3.88/4.07 Clause #27 (by eliminate duplicate literals #[26]): Eq b True % 3.88/4.07 Clause #32 (by clausification #[22]): Or (Eq (p True) False) (Eq b False) % 3.88/4.07 Clause #33 (by forward demodulation #[32, 27]): Or (Eq (p True) False) (Eq True False) % 3.88/4.07 Clause #34 (by clausification #[33]): Eq (p True) False % 3.88/4.07 Clause #35 (by superposition #[34, 25]): Eq False True % 3.88/4.07 Clause #36 (by clausification #[35]): False % 3.88/4.07 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------