TSTP Solution File: SEV397^5 by Duper---1.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SEV397^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n005.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 19:24:53 EDT 2023
% Result : Theorem 5.69s 5.99s
% Output : Proof 5.69s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SEV397^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.12 % Command : duper %s
% 0.13/0.34 % Computer : n005.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Thu Aug 24 02:28:53 EDT 2023
% 0.13/0.34 % CPUTime :
% 5.69/5.99 SZS status Theorem for theBenchmark.p
% 5.69/5.99 SZS output start Proof for theBenchmark.p
% 5.69/5.99 Clause #0 (by assumption #[]): Eq (Not (∀ (Xx : a), Iff (Or (And (cX Xx) (cY Xx)) (cZ Xx)) (And (Or (cX Xx) (cZ Xx)) (Or (cY Xx) (cZ Xx))))) True
% 5.69/5.99 Clause #1 (by clausification #[0]): Eq (∀ (Xx : a), Iff (Or (And (cX Xx) (cY Xx)) (cZ Xx)) (And (Or (cX Xx) (cZ Xx)) (Or (cY Xx) (cZ Xx)))) False
% 5.69/5.99 Clause #2 (by clausification #[1]): ∀ (a_1 : a),
% 5.69/5.99 Eq
% 5.69/5.99 (Not
% 5.69/5.99 (Iff (Or (And (cX (skS.0 0 a_1)) (cY (skS.0 0 a_1))) (cZ (skS.0 0 a_1)))
% 5.69/5.99 (And (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1))))))
% 5.69/5.99 True
% 5.69/5.99 Clause #3 (by clausification #[2]): ∀ (a_1 : a),
% 5.69/5.99 Eq
% 5.69/5.99 (Iff (Or (And (cX (skS.0 0 a_1)) (cY (skS.0 0 a_1))) (cZ (skS.0 0 a_1)))
% 5.69/5.99 (And (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1)))))
% 5.69/5.99 False
% 5.69/5.99 Clause #4 (by clausification #[3]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (Or (And (cX (skS.0 0 a_1)) (cY (skS.0 0 a_1))) (cZ (skS.0 0 a_1))) False)
% 5.69/5.99 (Eq (And (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1)))) False)
% 5.69/5.99 Clause #5 (by clausification #[3]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (Or (And (cX (skS.0 0 a_1)) (cY (skS.0 0 a_1))) (cZ (skS.0 0 a_1))) True)
% 5.69/5.99 (Eq (And (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1)))) True)
% 5.69/5.99 Clause #6 (by clausification #[4]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (And (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1)))) False)
% 5.69/5.99 (Eq (cZ (skS.0 0 a_1)) False)
% 5.69/5.99 Clause #7 (by clausification #[4]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (And (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1)))) False)
% 5.69/5.99 (Eq (And (cX (skS.0 0 a_1)) (cY (skS.0 0 a_1))) False)
% 5.69/5.99 Clause #8 (by clausification #[6]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (cZ (skS.0 0 a_1)) False)
% 5.69/5.99 (Or (Eq (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) False) (Eq (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) False))
% 5.69/5.99 Clause #9 (by clausification #[8]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (cZ (skS.0 0 a_1)) False)
% 5.69/5.99 (Or (Eq (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) False) (Eq (cZ (skS.0 0 a_1)) False))
% 5.69/5.99 Clause #11 (by clausification #[9]): ∀ (a_1 : a), Or (Eq (cZ (skS.0 0 a_1)) False) (Or (Eq (cZ (skS.0 0 a_1)) False) (Eq (cZ (skS.0 0 a_1)) False))
% 5.69/5.99 Clause #13 (by eliminate duplicate literals #[11]): ∀ (a_1 : a), Eq (cZ (skS.0 0 a_1)) False
% 5.69/5.99 Clause #21 (by clausification #[7]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (And (cX (skS.0 0 a_1)) (cY (skS.0 0 a_1))) False)
% 5.69/5.99 (Or (Eq (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) False) (Eq (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) False))
% 5.69/5.99 Clause #22 (by clausification #[21]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) False)
% 5.69/5.99 (Or (Eq (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) False)
% 5.69/5.99 (Or (Eq (cX (skS.0 0 a_1)) False) (Eq (cY (skS.0 0 a_1)) False)))
% 5.69/5.99 Clause #24 (by clausification #[22]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) False)
% 5.69/5.99 (Or (Eq (cX (skS.0 0 a_1)) False) (Or (Eq (cY (skS.0 0 a_1)) False) (Eq (cX (skS.0 0 a_1)) False)))
% 5.69/5.99 Clause #29 (by clausification #[5]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (And (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1)))) True)
% 5.69/5.99 (Or (Eq (And (cX (skS.0 0 a_1)) (cY (skS.0 0 a_1))) True) (Eq (cZ (skS.0 0 a_1)) True))
% 5.69/5.99 Clause #30 (by clausification #[29]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (And (cX (skS.0 0 a_1)) (cY (skS.0 0 a_1))) True)
% 5.69/5.99 (Or (Eq (cZ (skS.0 0 a_1)) True) (Eq (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) True))
% 5.69/5.99 Clause #31 (by clausification #[29]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (And (cX (skS.0 0 a_1)) (cY (skS.0 0 a_1))) True)
% 5.69/5.99 (Or (Eq (cZ (skS.0 0 a_1)) True) (Eq (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) True))
% 5.69/5.99 Clause #32 (by clausification #[30]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (cZ (skS.0 0 a_1)) True) (Or (Eq (Or (cY (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) True) (Eq (cY (skS.0 0 a_1)) True))
% 5.69/5.99 Clause #34 (by clausification #[32]): ∀ (a_1 : a),
% 5.69/5.99 Or (Eq (cZ (skS.0 0 a_1)) True)
% 5.69/5.99 (Or (Eq (cY (skS.0 0 a_1)) True) (Or (Eq (cY (skS.0 0 a_1)) True) (Eq (cZ (skS.0 0 a_1)) True)))
% 5.69/6.01 Clause #35 (by eliminate duplicate literals #[34]): ∀ (a_1 : a), Or (Eq (cZ (skS.0 0 a_1)) True) (Eq (cY (skS.0 0 a_1)) True)
% 5.69/6.01 Clause #36 (by forward demodulation #[35, 13]): ∀ (a_1 : a), Or (Eq False True) (Eq (cY (skS.0 0 a_1)) True)
% 5.69/6.01 Clause #37 (by clausification #[36]): ∀ (a_1 : a), Eq (cY (skS.0 0 a_1)) True
% 5.69/6.01 Clause #43 (by clausification #[24]): ∀ (a_1 : a),
% 5.69/6.01 Or (Eq (cX (skS.0 0 a_1)) False)
% 5.69/6.01 (Or (Eq (cY (skS.0 0 a_1)) False) (Or (Eq (cX (skS.0 0 a_1)) False) (Eq (cY (skS.0 0 a_1)) False)))
% 5.69/6.01 Clause #44 (by eliminate duplicate literals #[43]): ∀ (a_1 : a), Or (Eq (cX (skS.0 0 a_1)) False) (Eq (cY (skS.0 0 a_1)) False)
% 5.69/6.01 Clause #46 (by clausification #[31]): ∀ (a_1 : a),
% 5.69/6.01 Or (Eq (cZ (skS.0 0 a_1)) True) (Or (Eq (Or (cX (skS.0 0 a_1)) (cZ (skS.0 0 a_1))) True) (Eq (cX (skS.0 0 a_1)) True))
% 5.69/6.01 Clause #51 (by clausification #[46]): ∀ (a_1 : a),
% 5.69/6.01 Or (Eq (cZ (skS.0 0 a_1)) True)
% 5.69/6.01 (Or (Eq (cX (skS.0 0 a_1)) True) (Or (Eq (cX (skS.0 0 a_1)) True) (Eq (cZ (skS.0 0 a_1)) True)))
% 5.69/6.01 Clause #52 (by eliminate duplicate literals #[51]): ∀ (a_1 : a), Or (Eq (cZ (skS.0 0 a_1)) True) (Eq (cX (skS.0 0 a_1)) True)
% 5.69/6.01 Clause #53 (by forward demodulation #[52, 13]): ∀ (a_1 : a), Or (Eq False True) (Eq (cX (skS.0 0 a_1)) True)
% 5.69/6.01 Clause #54 (by clausification #[53]): ∀ (a_1 : a), Eq (cX (skS.0 0 a_1)) True
% 5.69/6.01 Clause #55 (by backward demodulation #[54, 44]): ∀ (a_1 : a), Or (Eq True False) (Eq (cY (skS.0 0 a_1)) False)
% 5.69/6.01 Clause #56 (by clausification #[55]): ∀ (a_1 : a), Eq (cY (skS.0 0 a_1)) False
% 5.69/6.01 Clause #57 (by superposition #[56, 37]): Eq False True
% 5.69/6.01 Clause #58 (by clausification #[57]): False
% 5.69/6.01 SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------