TSTP Solution File: SYN722+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN722+1 : TPTP v8.1.2. Released v2.5.0.
% 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 : n006.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 03:35:51 EDT 2023
% Result : Theorem 0.17s 0.39s
% Output : Proof 0.17s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SYN722+1 : TPTP v8.1.2. Released v2.5.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.17/0.35 % Computer : n006.cluster.edu
% 0.17/0.35 % Model : x86_64 x86_64
% 0.17/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.17/0.35 % Memory : 8042.1875MB
% 0.17/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.17/0.35 % CPULimit : 300
% 0.17/0.35 % WCLimit : 300
% 0.17/0.35 % DateTime : Sat Aug 26 17:27:07 EDT 2023
% 0.17/0.35 % CPUTime :
% 0.17/0.39 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 0.17/0.39
% 0.17/0.39 % SZS status Theorem
% 0.17/0.39
% 0.17/0.39 % SZS output start Proof
% 0.17/0.39 Take the following subset of the input axioms:
% 0.17/0.39 fof(thm119, conjecture, ~(![Z]: ((p(Z) | r(Z)) & q(Z)) & (![X]: ?[Y]: (p(X) | (~q(X) | (~q(Y) | (~q(c) | ~q(d))))) & (~p(a) | ~p(b))))).
% 0.17/0.39
% 0.17/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.17/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.17/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.17/0.39 fresh(y, y, x1...xn) = u
% 0.17/0.39 C => fresh(s, t, x1...xn) = v
% 0.17/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.17/0.39 variables of u and v.
% 0.17/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.17/0.39 input problem has no model of domain size 1).
% 0.17/0.39
% 0.17/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.17/0.39
% 0.17/0.39 Axiom 1 (thm119_1): q(X) = true.
% 0.17/0.39 Axiom 2 (thm119_3): fresh4(X, X, Y) = true.
% 0.17/0.39 Axiom 3 (thm119_3): fresh2(X, X, Y) = p(Y).
% 0.17/0.39 Axiom 4 (thm119_3): fresh3(X, X, Y) = fresh4(q(Y), true, Y).
% 0.17/0.39 Axiom 5 (thm119_3): fresh(q(y), true, X) = fresh3(q(d), true, X).
% 0.17/0.39 Axiom 6 (thm119_3): fresh(X, X, Y) = fresh2(q(c), true, Y).
% 0.17/0.39
% 0.17/0.39 Lemma 7: p(X) = true.
% 0.17/0.39 Proof:
% 0.17/0.39 p(X)
% 0.17/0.39 = { by axiom 3 (thm119_3) R->L }
% 0.17/0.39 fresh2(true, true, X)
% 0.17/0.39 = { by axiom 1 (thm119_1) R->L }
% 0.17/0.39 fresh2(q(c), true, X)
% 0.17/0.39 = { by axiom 6 (thm119_3) R->L }
% 0.17/0.39 fresh(true, true, X)
% 0.17/0.39 = { by axiom 1 (thm119_1) R->L }
% 0.17/0.39 fresh(q(y), true, X)
% 0.17/0.39 = { by axiom 5 (thm119_3) }
% 0.17/0.39 fresh3(q(d), true, X)
% 0.17/0.39 = { by axiom 1 (thm119_1) }
% 0.17/0.39 fresh3(true, true, X)
% 0.17/0.39 = { by axiom 4 (thm119_3) }
% 0.17/0.39 fresh4(q(X), true, X)
% 0.17/0.39 = { by axiom 1 (thm119_1) }
% 0.17/0.39 fresh4(true, true, X)
% 0.17/0.39 = { by axiom 2 (thm119_3) }
% 0.17/0.39 true
% 0.17/0.39
% 0.17/0.39 Goal 1 (thm119_2): tuple(p(a), p(b)) = tuple(true, true).
% 0.17/0.39 Proof:
% 0.17/0.39 tuple(p(a), p(b))
% 0.17/0.39 = { by lemma 7 }
% 0.17/0.39 tuple(p(a), true)
% 0.17/0.39 = { by lemma 7 }
% 0.17/0.39 tuple(true, true)
% 0.17/0.39 % SZS output end Proof
% 0.17/0.39
% 0.17/0.39 RESULT: Theorem (the conjecture is true).
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