TSTP Solution File: SYN379+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN379+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 : n007.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 03:34:22 EDT 2023
% Result : Theorem 0.13s 0.37s
% Output : Proof 0.13s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SYN379+1 : TPTP v8.1.2. Released v2.0.0.
% 0.07/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34 % Computer : n007.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 : Sat Aug 26 20:29:57 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.13/0.37 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.13/0.37
% 0.13/0.37 % SZS status Theorem
% 0.13/0.37
% 0.13/0.37 % SZS output start Proof
% 0.13/0.37 Take the following subset of the input axioms:
% 0.13/0.37 fof(x2131, conjecture, ![X]: big_p(X) => ?[Y]: (![Z, X2]: big_q(X2, Y, Z) => ~![Z2]: (big_p(Z2) & ~big_q(Y, Y, Z2)))).
% 0.13/0.37
% 0.13/0.37 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.13/0.37 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.13/0.37 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.13/0.37 fresh(y, y, x1...xn) = u
% 0.13/0.37 C => fresh(s, t, x1...xn) = v
% 0.13/0.37 where fresh is a fresh function symbol and x1..xn are the free
% 0.13/0.37 variables of u and v.
% 0.13/0.37 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.13/0.37 input problem has no model of domain size 1).
% 0.13/0.37
% 0.13/0.37 The encoding turns the above axioms into the following unit equations and goals:
% 0.13/0.37
% 0.13/0.37 Axiom 1 (x2131_2): big_q(X, Y, Z) = true2.
% 0.13/0.37
% 0.13/0.37 Goal 1 (x2131_3): big_q(X, X, Y) = true2.
% 0.13/0.37 The goal is true when:
% 0.13/0.38 X = X
% 0.13/0.38 Y = Y
% 0.13/0.38
% 0.13/0.38 Proof:
% 0.13/0.38 big_q(X, X, Y)
% 0.13/0.38 = { by axiom 1 (x2131_2) }
% 0.13/0.38 true2
% 0.13/0.38 % SZS output end Proof
% 0.13/0.38
% 0.13/0.38 RESULT: Theorem (the conjecture is true).
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