TSTP Solution File: SEU885^5 by Duper---1.0
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%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SEU885^5 : 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 3.80s 4.01s
% Output : Proof 3.80s
% 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 : SEU885^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14 % Command : duper %s
% 0.15/0.34 % Computer : n027.cluster.edu
% 0.15/0.34 % Model : x86_64 x86_64
% 0.15/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.34 % Memory : 8042.1875MB
% 0.15/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.15/0.34 % CPULimit : 300
% 0.15/0.34 % WCLimit : 300
% 0.15/0.34 % DateTime : Wed Aug 23 13:33:31 EDT 2023
% 0.15/0.34 % CPUTime :
% 3.80/4.01 SZS status Theorem for theBenchmark.p
% 3.80/4.01 SZS output start Proof for theBenchmark.p
% 3.80/4.01 Clause #0 (by assumption #[]): Eq
% 3.80/4.01 (Not
% 3.80/4.01 ((∀ (Xx : b), cU Xx → cV Xx) →
% 3.80/4.01 ∀ (Xx : a), (Exists fun Xt => And (cU Xt) (Eq Xx (cF Xt))) → Exists fun Xt => And (cV Xt) (Eq Xx (cF Xt))))
% 3.80/4.01 True
% 3.80/4.01 Clause #1 (by clausification #[0]): Eq
% 3.80/4.01 ((∀ (Xx : b), cU Xx → cV Xx) →
% 3.80/4.01 ∀ (Xx : a), (Exists fun Xt => And (cU Xt) (Eq Xx (cF Xt))) → Exists fun Xt => And (cV Xt) (Eq Xx (cF Xt)))
% 3.80/4.01 False
% 3.80/4.01 Clause #2 (by clausification #[1]): Eq (∀ (Xx : b), cU Xx → cV Xx) True
% 3.80/4.01 Clause #3 (by clausification #[1]): Eq (∀ (Xx : a), (Exists fun Xt => And (cU Xt) (Eq Xx (cF Xt))) → Exists fun Xt => And (cV Xt) (Eq Xx (cF Xt))) False
% 3.80/4.01 Clause #4 (by clausification #[2]): ∀ (a : b), Eq (cU a → cV a) True
% 3.80/4.01 Clause #5 (by clausification #[4]): ∀ (a : b), Or (Eq (cU a) False) (Eq (cV a) True)
% 3.80/4.01 Clause #6 (by clausification #[3]): ∀ (a_1 : a),
% 3.80/4.01 Eq
% 3.80/4.01 (Not
% 3.80/4.01 ((Exists fun Xt => And (cU Xt) (Eq (skS.0 0 a_1) (cF Xt))) →
% 3.80/4.01 Exists fun Xt => And (cV Xt) (Eq (skS.0 0 a_1) (cF Xt))))
% 3.80/4.01 True
% 3.80/4.01 Clause #7 (by clausification #[6]): ∀ (a_1 : a),
% 3.80/4.01 Eq
% 3.80/4.01 ((Exists fun Xt => And (cU Xt) (Eq (skS.0 0 a_1) (cF Xt))) →
% 3.80/4.01 Exists fun Xt => And (cV Xt) (Eq (skS.0 0 a_1) (cF Xt)))
% 3.80/4.01 False
% 3.80/4.01 Clause #8 (by clausification #[7]): ∀ (a_1 : a), Eq (Exists fun Xt => And (cU Xt) (Eq (skS.0 0 a_1) (cF Xt))) True
% 3.80/4.01 Clause #9 (by clausification #[7]): ∀ (a_1 : a), Eq (Exists fun Xt => And (cV Xt) (Eq (skS.0 0 a_1) (cF Xt))) False
% 3.80/4.01 Clause #10 (by clausification #[8]): ∀ (a_1 : a) (a_2 : b), Eq (And (cU (skS.0 1 a_1 a_2)) (Eq (skS.0 0 a_1) (cF (skS.0 1 a_1 a_2)))) True
% 3.80/4.01 Clause #11 (by clausification #[10]): ∀ (a_1 : a) (a_2 : b), Eq (Eq (skS.0 0 a_1) (cF (skS.0 1 a_1 a_2))) True
% 3.80/4.01 Clause #12 (by clausification #[10]): ∀ (a_1 : a) (a_2 : b), Eq (cU (skS.0 1 a_1 a_2)) True
% 3.80/4.01 Clause #13 (by clausification #[11]): ∀ (a_1 : a) (a_2 : b), Eq (skS.0 0 a_1) (cF (skS.0 1 a_1 a_2))
% 3.80/4.01 Clause #14 (by superposition #[12, 5]): ∀ (a_1 : a) (a_2 : b), Or (Eq True False) (Eq (cV (skS.0 1 a_1 a_2)) True)
% 3.80/4.01 Clause #15 (by clausification #[14]): ∀ (a_1 : a) (a_2 : b), Eq (cV (skS.0 1 a_1 a_2)) True
% 3.80/4.01 Clause #16 (by clausification #[9]): ∀ (a_1 : b) (a_2 : a), Eq (And (cV a_1) (Eq (skS.0 0 a_2) (cF a_1))) False
% 3.80/4.01 Clause #17 (by clausification #[16]): ∀ (a_1 : b) (a_2 : a), Or (Eq (cV a_1) False) (Eq (Eq (skS.0 0 a_2) (cF a_1)) False)
% 3.80/4.01 Clause #18 (by clausification #[17]): ∀ (a_1 : b) (a_2 : a), Or (Eq (cV a_1) False) (Ne (skS.0 0 a_2) (cF a_1))
% 3.80/4.01 Clause #19 (by superposition #[18, 15]): ∀ (a_1 a_2 : a) (a_3 : b), Or (Ne (skS.0 0 a_1) (cF (skS.0 1 a_2 a_3))) (Eq False True)
% 3.80/4.01 Clause #20 (by clausification #[19]): ∀ (a_1 a_2 : a) (a_3 : b), Ne (skS.0 0 a_1) (cF (skS.0 1 a_2 a_3))
% 3.80/4.01 Clause #21 (by forward demodulation #[20, 13]): ∀ (a_1 a_2 : a), Ne (skS.0 0 a_1) (skS.0 0 a_2)
% 3.80/4.01 Clause #22 (by equality resolution #[21]): False
% 3.80/4.01 SZS output end Proof for theBenchmark.p
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