TSTP Solution File: SYN359+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN359+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 : n023.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 03:34:17 EDT 2023
% Result : Theorem 0.13s 0.38s
% 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 : SYN359+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 : n023.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 17:59:41 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.13/0.38 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.13/0.38
% 0.13/0.38 % SZS status Theorem
% 0.13/0.38
% 0.13/0.38 % SZS output start Proof
% 0.13/0.38 Take the following subset of the input axioms:
% 0.13/0.38 fof(x2110, conjecture, (?[X]: big_r(X) & (![Y]: (big_r(Y) => ?[Z]: big_q(Y, Z)) & ![X2, Y2]: (big_q(X2, Y2) => big_q(X2, X2)))) => ?[X2, Y2]: (big_q(X2, Y2) & big_r(Y2))).
% 0.13/0.38
% 0.13/0.38 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.13/0.38 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.13/0.38 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.13/0.38 fresh(y, y, x1...xn) = u
% 0.13/0.38 C => fresh(s, t, x1...xn) = v
% 0.13/0.38 where fresh is a fresh function symbol and x1..xn are the free
% 0.13/0.38 variables of u and v.
% 0.13/0.38 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.13/0.38 input problem has no model of domain size 1).
% 0.13/0.38
% 0.13/0.38 The encoding turns the above axioms into the following unit equations and goals:
% 0.13/0.38
% 0.13/0.39 Axiom 1 (x2110): big_r(x) = true2.
% 0.13/0.39 Axiom 2 (x2110_1): fresh(X, X, Y) = true2.
% 0.13/0.39 Axiom 3 (x2110_3): fresh2(X, X, Y) = true2.
% 0.13/0.39 Axiom 4 (x2110_1): fresh(big_r(X), true2, X) = big_q(X, z(X)).
% 0.13/0.39 Axiom 5 (x2110_3): fresh2(big_q(X, Y), true2, X) = big_q(X, X).
% 0.13/0.39
% 0.13/0.39 Goal 1 (x2110_2): tuple(big_r(X), big_q(Y, X)) = tuple(true2, true2).
% 0.13/0.39 The goal is true when:
% 0.13/0.39 X = x
% 0.13/0.39 Y = x
% 0.13/0.39
% 0.13/0.39 Proof:
% 0.13/0.39 tuple(big_r(x), big_q(x, x))
% 0.13/0.39 = { by axiom 5 (x2110_3) R->L }
% 0.13/0.39 tuple(big_r(x), fresh2(big_q(x, z(x)), true2, x))
% 0.13/0.39 = { by axiom 4 (x2110_1) R->L }
% 0.13/0.39 tuple(big_r(x), fresh2(fresh(big_r(x), true2, x), true2, x))
% 0.13/0.39 = { by axiom 1 (x2110) }
% 0.13/0.39 tuple(big_r(x), fresh2(fresh(true2, true2, x), true2, x))
% 0.13/0.39 = { by axiom 2 (x2110_1) }
% 0.13/0.39 tuple(big_r(x), fresh2(true2, true2, x))
% 0.13/0.39 = { by axiom 3 (x2110_3) }
% 0.13/0.39 tuple(big_r(x), true2)
% 0.13/0.39 = { by axiom 1 (x2110) }
% 0.13/0.39 tuple(true2, true2)
% 0.13/0.39 % SZS output end Proof
% 0.13/0.39
% 0.13/0.39 RESULT: Theorem (the conjecture is true).
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