TSTP Solution File: SEU886^5 by Duper---1.0
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- Process Solution
%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SEU886^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n027.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 16:44:13 EDT 2023
% Result : Theorem 3.79s 3.95s
% Output : Proof 3.79s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SEU886^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14 % Command : duper %s
% 0.13/0.35 % Computer : n027.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Wed Aug 23 14:27:01 EDT 2023
% 0.13/0.35 % CPUTime :
% 3.79/3.95 SZS status Theorem for theBenchmark.p
% 3.79/3.95 SZS output start Proof for theBenchmark.p
% 3.79/3.95 Clause #0 (by assumption #[]): Eq
% 3.79/3.95 (Not
% 3.79/3.95 (∀ (Xx0 : a),
% 3.79/3.95 (Exists fun Xt => And (And (x Xt) (y Xt)) (Eq Xx0 (f Xt))) →
% 3.79/3.95 And (Exists fun Xt => And (x Xt) (Eq Xx0 (f Xt))) (Exists fun Xt => And (y Xt) (Eq Xx0 (f Xt)))))
% 3.79/3.95 True
% 3.79/3.95 Clause #1 (by clausification #[0]): Eq
% 3.79/3.95 (∀ (Xx0 : a),
% 3.79/3.95 (Exists fun Xt => And (And (x Xt) (y Xt)) (Eq Xx0 (f Xt))) →
% 3.79/3.95 And (Exists fun Xt => And (x Xt) (Eq Xx0 (f Xt))) (Exists fun Xt => And (y Xt) (Eq Xx0 (f Xt))))
% 3.79/3.95 False
% 3.79/3.95 Clause #2 (by clausification #[1]): ∀ (a_1 : a),
% 3.79/3.95 Eq
% 3.79/3.95 (Not
% 3.79/3.95 ((Exists fun Xt => And (And (x Xt) (y Xt)) (Eq (skS.0 0 a_1) (f Xt))) →
% 3.79/3.95 And (Exists fun Xt => And (x Xt) (Eq (skS.0 0 a_1) (f Xt)))
% 3.79/3.95 (Exists fun Xt => And (y Xt) (Eq (skS.0 0 a_1) (f Xt)))))
% 3.79/3.95 True
% 3.79/3.95 Clause #3 (by clausification #[2]): ∀ (a_1 : a),
% 3.79/3.95 Eq
% 3.79/3.95 ((Exists fun Xt => And (And (x Xt) (y Xt)) (Eq (skS.0 0 a_1) (f Xt))) →
% 3.79/3.95 And (Exists fun Xt => And (x Xt) (Eq (skS.0 0 a_1) (f Xt)))
% 3.79/3.95 (Exists fun Xt => And (y Xt) (Eq (skS.0 0 a_1) (f Xt))))
% 3.79/3.95 False
% 3.79/3.95 Clause #4 (by clausification #[3]): ∀ (a_1 : a), Eq (Exists fun Xt => And (And (x Xt) (y Xt)) (Eq (skS.0 0 a_1) (f Xt))) True
% 3.79/3.95 Clause #5 (by clausification #[3]): ∀ (a_1 : a),
% 3.79/3.95 Eq
% 3.79/3.95 (And (Exists fun Xt => And (x Xt) (Eq (skS.0 0 a_1) (f Xt)))
% 3.79/3.95 (Exists fun Xt => And (y Xt) (Eq (skS.0 0 a_1) (f Xt))))
% 3.79/3.95 False
% 3.79/3.95 Clause #6 (by clausification #[4]): ∀ (a_1 : a) (a_2 : b),
% 3.79/3.95 Eq (And (And (x (skS.0 1 a_1 a_2)) (y (skS.0 1 a_1 a_2))) (Eq (skS.0 0 a_1) (f (skS.0 1 a_1 a_2)))) True
% 3.79/3.95 Clause #7 (by clausification #[6]): ∀ (a_1 : a) (a_2 : b), Eq (Eq (skS.0 0 a_1) (f (skS.0 1 a_1 a_2))) True
% 3.79/3.95 Clause #8 (by clausification #[6]): ∀ (a_1 : a) (a_2 : b), Eq (And (x (skS.0 1 a_1 a_2)) (y (skS.0 1 a_1 a_2))) True
% 3.79/3.95 Clause #9 (by clausification #[7]): ∀ (a_1 : a) (a_2 : b), Eq (skS.0 0 a_1) (f (skS.0 1 a_1 a_2))
% 3.79/3.95 Clause #10 (by clausification #[8]): ∀ (a_1 : a) (a_2 : b), Eq (y (skS.0 1 a_1 a_2)) True
% 3.79/3.95 Clause #11 (by clausification #[8]): ∀ (a_1 : a) (a_2 : b), Eq (x (skS.0 1 a_1 a_2)) True
% 3.79/3.95 Clause #12 (by clausification #[5]): ∀ (a_1 : a),
% 3.79/3.95 Or (Eq (Exists fun Xt => And (x Xt) (Eq (skS.0 0 a_1) (f Xt))) False)
% 3.79/3.95 (Eq (Exists fun Xt => And (y Xt) (Eq (skS.0 0 a_1) (f Xt))) False)
% 3.79/3.95 Clause #13 (by clausification #[12]): ∀ (a_1 : a) (a_2 : b),
% 3.79/3.95 Or (Eq (Exists fun Xt => And (y Xt) (Eq (skS.0 0 a_1) (f Xt))) False)
% 3.79/3.95 (Eq (And (x a_2) (Eq (skS.0 0 a_1) (f a_2))) False)
% 3.79/3.95 Clause #14 (by clausification #[13]): ∀ (a_1 : b) (a_2 : a) (a_3 : b),
% 3.79/3.95 Or (Eq (And (x a_1) (Eq (skS.0 0 a_2) (f a_1))) False) (Eq (And (y a_3) (Eq (skS.0 0 a_2) (f a_3))) False)
% 3.79/3.95 Clause #15 (by clausification #[14]): ∀ (a_1 : b) (a_2 : a) (a_3 : b),
% 3.79/3.95 Or (Eq (And (y a_1) (Eq (skS.0 0 a_2) (f a_1))) False) (Or (Eq (x a_3) False) (Eq (Eq (skS.0 0 a_2) (f a_3)) False))
% 3.79/3.95 Clause #16 (by clausification #[15]): ∀ (a_1 : b) (a_2 : a) (a_3 : b),
% 3.79/3.95 Or (Eq (x a_1) False)
% 3.79/3.95 (Or (Eq (Eq (skS.0 0 a_2) (f a_1)) False) (Or (Eq (y a_3) False) (Eq (Eq (skS.0 0 a_2) (f a_3)) False)))
% 3.79/3.95 Clause #17 (by clausification #[16]): ∀ (a_1 a_2 : b) (a_3 : a),
% 3.79/3.95 Or (Eq (x a_1) False) (Or (Eq (y a_2) False) (Or (Eq (Eq (skS.0 0 a_3) (f a_2)) False) (Ne (skS.0 0 a_3) (f a_1))))
% 3.79/3.95 Clause #18 (by clausification #[17]): ∀ (a_1 a_2 : b) (a_3 : a),
% 3.79/3.95 Or (Eq (x a_1) False) (Or (Eq (y a_2) False) (Or (Ne (skS.0 0 a_3) (f a_1)) (Ne (skS.0 0 a_3) (f a_2))))
% 3.79/3.95 Clause #19 (by superposition #[18, 11]): ∀ (a_1 : b) (a_2 a_3 : a) (a_4 : b),
% 3.79/3.95 Or (Eq (y a_1) False) (Or (Ne (skS.0 0 a_2) (f (skS.0 1 a_3 a_4))) (Or (Ne (skS.0 0 a_2) (f a_1)) (Eq False True)))
% 3.79/3.95 Clause #20 (by clausification #[19]): ∀ (a_1 : b) (a_2 a_3 : a) (a_4 : b),
% 3.79/3.95 Or (Eq (y a_1) False) (Or (Ne (skS.0 0 a_2) (f (skS.0 1 a_3 a_4))) (Ne (skS.0 0 a_2) (f a_1)))
% 3.79/3.95 Clause #21 (by forward demodulation #[20, 9]): ∀ (a_1 : b) (a_2 a_3 : a), Or (Eq (y a_1) False) (Or (Ne (skS.0 0 a_2) (skS.0 0 a_3)) (Ne (skS.0 0 a_2) (f a_1)))
% 3.79/3.95 Clause #22 (by superposition #[21, 10]): ∀ (a_1 a_2 a_3 : a) (a_4 : b),
% 3.79/3.95 Or (Ne (skS.0 0 a_1) (skS.0 0 a_2)) (Or (Ne (skS.0 0 a_1) (f (skS.0 1 a_3 a_4))) (Eq False True))
% 3.79/3.95 Clause #23 (by clausification #[22]): ∀ (a_1 a_2 a_3 : a) (a_4 : b), Or (Ne (skS.0 0 a_1) (skS.0 0 a_2)) (Ne (skS.0 0 a_1) (f (skS.0 1 a_3 a_4)))
% 3.79/3.95 Clause #24 (by forward demodulation #[23, 9]): ∀ (a_1 a_2 a_3 : a), Or (Ne (skS.0 0 a_1) (skS.0 0 a_2)) (Ne (skS.0 0 a_1) (skS.0 0 a_3))
% 3.79/3.95 Clause #25 (by equality resolution #[24]): ∀ (a_1 a_2 : a), Ne (skS.0 0 a_1) (skS.0 0 a_2)
% 3.79/3.95 Clause #26 (by equality resolution #[25]): False
% 3.79/3.95 SZS output end Proof for theBenchmark.p
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