TSTP Solution File: SWC041+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWC041+1 : TPTP v8.1.2. Released v2.4.0.
% 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 : n025.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 : Thu Aug 31 20:53:34 EDT 2023
% Result : Theorem 4.25s 0.90s
% Output : Proof 4.25s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.11 % Problem : SWC041+1 : TPTP v8.1.2. Released v2.4.0.
% 0.00/0.12 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.33 % Computer : n025.cluster.edu
% 0.14/0.33 % Model : x86_64 x86_64
% 0.14/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.33 % Memory : 8042.1875MB
% 0.14/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.33 % CPULimit : 300
% 0.14/0.33 % WCLimit : 300
% 0.14/0.33 % DateTime : Mon Aug 28 17:28:54 EDT 2023
% 0.14/0.33 % CPUTime :
% 4.25/0.90 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 4.25/0.90
% 4.25/0.90 % SZS status Theorem
% 4.25/0.90
% 4.25/0.90 % SZS output start Proof
% 4.25/0.90 Take the following subset of the input axioms:
% 4.25/0.91 fof(ax83, axiom, ![U]: (ssList(U) => ![V]: (ssList(V) => (nil=app(U, V) <=> (nil=V & nil=U))))).
% 4.25/0.91 fof(co1, conjecture, ![U2]: (ssList(U2) => ![V2]: (ssList(V2) => ![W]: (ssList(W) => ![X]: (ssList(X) => (nil!=V2 | (V2!=X | (U2!=W | (nil=U2 | (![Y]: (ssList(Y) => (app(W, Y)!=X | (~equalelemsP(W) | ?[Z]: (ssItem(Z) & ?[X1]: (ssList(X1) & (app(cons(Z, nil), X1)=Y & ?[X2]: (ssList(X2) & app(X2, cons(Z, nil))=W))))))) | (nil!=X & nil=W))))))))))).
% 4.25/0.91
% 4.25/0.91 Now clausify the problem and encode Horn clauses using encoding 3 of
% 4.25/0.91 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 4.25/0.91 We repeatedly replace C & s=t => u=v by the two clauses:
% 4.25/0.91 fresh(y, y, x1...xn) = u
% 4.25/0.91 C => fresh(s, t, x1...xn) = v
% 4.25/0.91 where fresh is a fresh function symbol and x1..xn are the free
% 4.25/0.91 variables of u and v.
% 4.25/0.91 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 4.25/0.91 input problem has no model of domain size 1).
% 4.25/0.91
% 4.25/0.91 The encoding turns the above axioms into the following unit equations and goals:
% 4.25/0.91
% 4.25/0.91 Axiom 1 (co1_2): u = w.
% 4.25/0.91 Axiom 2 (co1_1): nil = v.
% 4.25/0.91 Axiom 3 (co1_3): v = x.
% 4.25/0.91 Axiom 4 (co1_6): ssList(w) = true2.
% 4.25/0.91 Axiom 5 (co1_8): ssList(y) = true2.
% 4.25/0.91 Axiom 6 (co1): app(w, y) = x.
% 4.25/0.91 Axiom 7 (ax83_1): fresh116(X, X, Y) = Y.
% 4.25/0.91 Axiom 8 (ax83_1): fresh22(X, X, Y, Z) = nil.
% 4.25/0.91 Axiom 9 (ax83_1): fresh115(X, X, Y, Z) = fresh116(nil, app(Y, Z), Y).
% 4.25/0.91 Axiom 10 (ax83_1): fresh115(ssList(X), true2, Y, X) = fresh22(ssList(Y), true2, Y, X).
% 4.25/0.91
% 4.25/0.91 Goal 1 (co1_11): nil = u.
% 4.25/0.91 Proof:
% 4.25/0.91 nil
% 4.25/0.91 = { by axiom 8 (ax83_1) R->L }
% 4.25/0.91 fresh22(true2, true2, w, y)
% 4.25/0.91 = { by axiom 4 (co1_6) R->L }
% 4.25/0.91 fresh22(ssList(w), true2, w, y)
% 4.25/0.91 = { by axiom 10 (ax83_1) R->L }
% 4.25/0.91 fresh115(ssList(y), true2, w, y)
% 4.25/0.91 = { by axiom 5 (co1_8) }
% 4.25/0.91 fresh115(true2, true2, w, y)
% 4.25/0.91 = { by axiom 9 (ax83_1) }
% 4.25/0.91 fresh116(nil, app(w, y), w)
% 4.25/0.91 = { by axiom 6 (co1) }
% 4.25/0.91 fresh116(nil, x, w)
% 4.25/0.91 = { by axiom 3 (co1_3) R->L }
% 4.25/0.91 fresh116(nil, v, w)
% 4.25/0.91 = { by axiom 2 (co1_1) R->L }
% 4.25/0.91 fresh116(nil, nil, w)
% 4.25/0.91 = { by axiom 7 (ax83_1) }
% 4.25/0.91 w
% 4.25/0.91 = { by axiom 1 (co1_2) R->L }
% 4.25/0.91 u
% 4.25/0.91 % SZS output end Proof
% 4.25/0.91
% 4.25/0.91 RESULT: Theorem (the conjecture is true).
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