TSTP Solution File: SYN064+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SYN064+1 : TPTP v8.1.2. Released v2.0.0.
% Transfm : none
% Format : tptp:raw
% Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% Computer : n028.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 Sep 1 03:32:59 EDT 2023
% Result : Theorem 0.21s 0.40s
% Output : Proof 0.21s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.14 % Problem : SYN064+1 : TPTP v8.1.2. Released v2.0.0.
% 0.00/0.15 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.36 % Computer : n028.cluster.edu
% 0.14/0.36 % Model : x86_64 x86_64
% 0.14/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.36 % Memory : 8042.1875MB
% 0.14/0.36 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.36 % CPULimit : 300
% 0.14/0.36 % WCLimit : 300
% 0.14/0.36 % DateTime : Sat Aug 26 21:16:22 EDT 2023
% 0.14/0.37 % CPUTime :
% 0.21/0.40 Command-line arguments: --no-flatten-goal
% 0.21/0.40
% 0.21/0.40 % SZS status Theorem
% 0.21/0.40
% 0.21/0.40 % SZS output start Proof
% 0.21/0.40 Take the following subset of the input axioms:
% 0.21/0.40 fof(pel35, conjecture, ?[X, Y]: (big_p(X, Y) => ![Z, W]: big_p(Z, W))).
% 0.21/0.40
% 0.21/0.40 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.40 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.40 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.40 fresh(y, y, x1...xn) = u
% 0.21/0.40 C => fresh(s, t, x1...xn) = v
% 0.21/0.40 where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.40 variables of u and v.
% 0.21/0.40 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.40 input problem has no model of domain size 1).
% 0.21/0.40
% 0.21/0.40 The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.40
% 0.21/0.40 Axiom 1 (pel35): big_p(X, Y) = true.
% 0.21/0.40
% 0.21/0.40 Goal 1 (pel35_1): big_p(z, w) = true.
% 0.21/0.40 Proof:
% 0.21/0.40 big_p(z, w)
% 0.21/0.40 = { by axiom 1 (pel35) }
% 0.21/0.40 true
% 0.21/0.40 % SZS output end Proof
% 0.21/0.40
% 0.21/0.40 RESULT: Theorem (the conjecture is true).
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