TSTP Solution File: SYN068+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN068+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 : n021.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 03:33:01 EDT 2023
% Result : Theorem 0.19s 0.39s
% Output : Proof 0.19s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : SYN068+1 : TPTP v8.1.2. Released v2.0.0.
% 0.12/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.35 % Computer : n021.cluster.edu
% 0.14/0.35 % Model : x86_64 x86_64
% 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35 % Memory : 8042.1875MB
% 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35 % CPULimit : 300
% 0.14/0.35 % WCLimit : 300
% 0.14/0.35 % DateTime : Sat Aug 26 19:54:42 EDT 2023
% 0.14/0.35 % CPUTime :
% 0.19/0.39 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.19/0.39
% 0.19/0.39 % SZS status Theorem
% 0.19/0.39
% 0.19/0.39 % SZS output start Proof
% 0.19/0.39 Take the following subset of the input axioms:
% 0.19/0.39 fof(pel44, conjecture, ?[X]: (big_j(X) & ~big_f(X))).
% 0.19/0.39 fof(pel44_1, axiom, ![X2]: (big_f(X2) => (?[Y]: (big_g(Y) & big_h(X2, Y)) & ?[Y1]: (big_g(Y1) & ~big_h(X2, Y1))))).
% 0.19/0.39 fof(pel44_2, axiom, ?[X2]: (big_j(X2) & ![Y2]: (big_g(Y2) => big_h(X2, Y2)))).
% 0.19/0.39
% 0.19/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.39 fresh(y, y, x1...xn) = u
% 0.19/0.39 C => fresh(s, t, x1...xn) = v
% 0.19/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.39 variables of u and v.
% 0.19/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.39 input problem has no model of domain size 1).
% 0.19/0.39
% 0.19/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.39
% 0.19/0.39 Axiom 1 (pel44_2): big_j(x) = true2.
% 0.19/0.39 Axiom 2 (pel44_2_1): fresh(X, X, Y) = true2.
% 0.19/0.39 Axiom 3 (pel44): fresh4(X, X, Y) = true2.
% 0.19/0.39 Axiom 4 (pel44_1_1): fresh3(X, X, Y) = true2.
% 0.19/0.39 Axiom 5 (pel44_2_1): fresh(big_g(X), true2, X) = big_h(x, X).
% 0.19/0.39 Axiom 6 (pel44): fresh4(big_j(X), true2, X) = big_f(X).
% 0.19/0.39 Axiom 7 (pel44_1_1): fresh3(big_f(X), true2, X) = big_g(y1(X)).
% 0.19/0.39
% 0.19/0.39 Lemma 8: big_f(x) = true2.
% 0.19/0.39 Proof:
% 0.19/0.39 big_f(x)
% 0.19/0.39 = { by axiom 6 (pel44) R->L }
% 0.19/0.39 fresh4(big_j(x), true2, x)
% 0.19/0.39 = { by axiom 1 (pel44_2) }
% 0.19/0.39 fresh4(true2, true2, x)
% 0.19/0.39 = { by axiom 3 (pel44) }
% 0.19/0.39 true2
% 0.19/0.39
% 0.19/0.40 Goal 1 (pel44_1_3): tuple(big_f(X), big_h(X, y1(X))) = tuple(true2, true2).
% 0.19/0.40 The goal is true when:
% 0.19/0.40 X = x
% 0.19/0.40
% 0.19/0.40 Proof:
% 0.19/0.40 tuple(big_f(x), big_h(x, y1(x)))
% 0.19/0.40 = { by axiom 5 (pel44_2_1) R->L }
% 0.19/0.40 tuple(big_f(x), fresh(big_g(y1(x)), true2, y1(x)))
% 0.19/0.40 = { by axiom 7 (pel44_1_1) R->L }
% 0.19/0.40 tuple(big_f(x), fresh(fresh3(big_f(x), true2, x), true2, y1(x)))
% 0.19/0.40 = { by lemma 8 }
% 0.19/0.40 tuple(big_f(x), fresh(fresh3(true2, true2, x), true2, y1(x)))
% 0.19/0.40 = { by axiom 4 (pel44_1_1) }
% 0.19/0.40 tuple(big_f(x), fresh(true2, true2, y1(x)))
% 0.19/0.40 = { by axiom 2 (pel44_2_1) }
% 0.19/0.40 tuple(big_f(x), true2)
% 0.19/0.40 = { by lemma 8 }
% 0.19/0.40 tuple(true2, true2)
% 0.19/0.40 % SZS output end Proof
% 0.19/0.40
% 0.19/0.40 RESULT: Theorem (the conjecture is true).
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