TSTP Solution File: SYN048+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN048+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 : n032.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 03:32:51 EDT 2023
% Result : Theorem 0.07s 0.30s
% Output : Proof 0.07s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.08 % Problem : SYN048+1 : TPTP v8.1.2. Released v2.0.0.
% 0.00/0.09 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.07/0.28 % Computer : n032.cluster.edu
% 0.07/0.28 % Model : x86_64 x86_64
% 0.07/0.28 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.07/0.28 % Memory : 8042.1875MB
% 0.07/0.28 % OS : Linux 3.10.0-693.el7.x86_64
% 0.07/0.28 % CPULimit : 300
% 0.07/0.28 % WCLimit : 300
% 0.07/0.28 % DateTime : Sat Aug 26 21:22:14 EDT 2023
% 0.07/0.28 % CPUTime :
% 0.07/0.30 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.07/0.30
% 0.07/0.30 % SZS status Theorem
% 0.07/0.30
% 0.07/0.30 % SZS output start Proof
% 0.07/0.30 Take the following subset of the input axioms:
% 0.07/0.30 fof(pel18, conjecture, ?[Y]: ![X]: (big_f(Y) => big_f(X))).
% 0.07/0.30
% 0.07/0.30 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.07/0.30 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.07/0.30 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.07/0.30 fresh(y, y, x1...xn) = u
% 0.07/0.30 C => fresh(s, t, x1...xn) = v
% 0.07/0.30 where fresh is a fresh function symbol and x1..xn are the free
% 0.07/0.30 variables of u and v.
% 0.07/0.30 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.07/0.30 input problem has no model of domain size 1).
% 0.07/0.30
% 0.07/0.30 The encoding turns the above axioms into the following unit equations and goals:
% 0.07/0.30
% 0.07/0.30 Axiom 1 (pel18): big_f(X) = true.
% 0.07/0.30
% 0.07/0.30 Goal 1 (pel18_1): big_f(x) = true.
% 0.07/0.30 Proof:
% 0.07/0.30 big_f(x)
% 0.07/0.30 = { by axiom 1 (pel18) }
% 0.07/0.30 true
% 0.07/0.30 % SZS output end Proof
% 0.07/0.30
% 0.07/0.30 RESULT: Theorem (the conjecture is true).
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