TSTP Solution File: SWC380+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWC380+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 : n027.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 20:55:16 EDT 2023
% Result : Theorem 5.63s 1.09s
% Output : Proof 5.63s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SWC380+1 : TPTP v8.1.2. Released v2.4.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.34 % Computer : n027.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 300
% 0.12/0.34 % DateTime : Mon Aug 28 15:31:37 EDT 2023
% 0.12/0.34 % CPUTime :
% 5.63/1.09 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 5.63/1.09
% 5.63/1.09 % SZS status Theorem
% 5.63/1.09
% 5.63/1.09 % SZS output start Proof
% 5.63/1.09 Take the following subset of the input axioms:
% 5.63/1.10 fof(ax1, axiom, ![U]: (ssItem(U) => ![V]: (ssItem(V) => (neq(U, V) <=> U!=V)))).
% 5.63/1.10 fof(ax13, axiom, ![U2]: (ssList(U2) => (duplicatefreeP(U2) <=> ![V2]: (ssItem(V2) => ![W]: (ssItem(W) => ![X]: (ssList(X) => ![Y]: (ssList(Y) => ![Z]: (ssList(Z) => (app(app(X, cons(V2, Y)), cons(W, Z))=U2 => V2!=W))))))))).
% 5.63/1.10 fof(ax15, axiom, ![U2]: (ssList(U2) => ![V2]: (ssList(V2) => (neq(U2, V2) <=> U2!=V2)))).
% 5.63/1.10 fof(ax18, axiom, ![U2]: (ssList(U2) => ![V2]: (ssItem(V2) => cons(V2, U2)!=U2))).
% 5.63/1.10 fof(ax21, axiom, ![U2]: (ssList(U2) => ![V2]: (ssItem(V2) => nil!=cons(V2, U2)))).
% 5.63/1.10 fof(ax33, axiom, ![U2]: (ssItem(U2) => ![V2]: (ssItem(V2) => (lt(U2, V2) => ~lt(V2, U2))))).
% 5.63/1.10 fof(ax37, axiom, ![U2]: (ssItem(U2) => ![V2]: (ssItem(V2) => ![W2]: (ssList(W2) => (memberP(cons(V2, W2), U2) <=> (U2=V2 | memberP(W2, U2))))))).
% 5.63/1.10 fof(ax38, axiom, ![U2]: (ssItem(U2) => ~memberP(nil, U2))).
% 5.63/1.10 fof(ax8, axiom, ![U2]: (ssList(U2) => (cyclefreeP(U2) <=> ![V2]: (ssItem(V2) => ![W2]: (ssItem(W2) => ![X2]: (ssList(X2) => ![Y2]: (ssList(Y2) => ![Z2]: (ssList(Z2) => (app(app(X2, cons(V2, Y2)), cons(W2, Z2))=U2 => ~(leq(V2, W2) & leq(W2, V2))))))))))).
% 5.63/1.10 fof(ax81, axiom, ![U2]: (ssList(U2) => ![V2]: (ssItem(V2) => cons(V2, U2)=app(cons(V2, nil), U2)))).
% 5.63/1.10 fof(ax90, axiom, ![U2]: (ssItem(U2) => ~lt(U2, U2))).
% 5.63/1.10 fof(ax93, axiom, ![U2]: (ssItem(U2) => ![V2]: (ssItem(V2) => (lt(U2, V2) <=> (U2!=V2 & leq(U2, V2)))))).
% 5.63/1.10 fof(ax94, axiom, ![U2]: (ssItem(U2) => ![V2]: (ssItem(V2) => (gt(U2, V2) => ~gt(V2, U2))))).
% 5.63/1.10 fof(co1, conjecture, ![U2]: (ssList(U2) => ![V2]: (ssList(V2) => ![W2]: (ssList(W2) => ![X2]: (ssList(X2) => (V2!=X2 | (U2!=W2 | ((~neq(V2, nil) | (?[Y2]: (ssItem(Y2) & (cons(Y2, nil)=U2 & memberP(V2, Y2))) | ![Z2]: (ssItem(Z2) => ![X1]: (ssList(X1) => (cons(Z2, nil)!=W2 | app(cons(Z2, nil), X1)!=X2))))) & (~neq(V2, nil) | neq(X2, nil)))))))))).
% 5.63/1.10
% 5.63/1.10 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.63/1.10 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.63/1.10 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.63/1.10 fresh(y, y, x1...xn) = u
% 5.63/1.10 C => fresh(s, t, x1...xn) = v
% 5.63/1.10 where fresh is a fresh function symbol and x1..xn are the free
% 5.63/1.10 variables of u and v.
% 5.63/1.10 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.63/1.10 input problem has no model of domain size 1).
% 5.63/1.10
% 5.63/1.10 The encoding turns the above axioms into the following unit equations and goals:
% 5.63/1.10
% 5.63/1.10 Axiom 1 (co1_3): v = x.
% 5.63/1.10 Axiom 2 (co1_2): u = w.
% 5.63/1.10 Axiom 3 (co1_5): neq(v, nil) = true2.
% 5.63/1.10 Axiom 4 (co1_13): fresh18(X, X) = x.
% 5.63/1.10 Axiom 5 (co1_14): fresh17(X, X) = w.
% 5.63/1.10 Axiom 6 (co1_15): fresh16(X, X) = true2.
% 5.63/1.10 Axiom 7 (co1_17): fresh14(X, X) = true2.
% 5.63/1.10 Axiom 8 (ax37_2): fresh69(X, X, Y, Z) = memberP(cons(Y, Z), Y).
% 5.63/1.10 Axiom 9 (ax37_2): fresh68(X, X, Y, Z) = true2.
% 5.63/1.10 Axiom 10 (ax81): fresh30(X, X, Y, Z) = app(cons(Z, nil), Y).
% 5.63/1.10 Axiom 11 (ax81): fresh29(X, X, Y, Z) = cons(Z, Y).
% 5.63/1.10 Axiom 12 (co1_13): fresh18(neq(x, nil), true2) = app(cons(z, nil), x1).
% 5.63/1.10 Axiom 13 (co1_14): fresh17(neq(x, nil), true2) = cons(z, nil).
% 5.63/1.10 Axiom 14 (co1_15): fresh16(neq(x, nil), true2) = ssItem(z).
% 5.63/1.10 Axiom 15 (co1_17): fresh14(neq(x, nil), true2) = ssList(x1).
% 5.63/1.10 Axiom 16 (ax37_2): fresh69(ssList(X), true2, Y, X) = fresh68(ssItem(Y), true2, Y, X).
% 5.63/1.10 Axiom 17 (ax81): fresh30(ssList(X), true2, X, Y) = fresh29(ssItem(Y), true2, X, Y).
% 5.63/1.10
% 5.63/1.10 Lemma 18: neq(x, nil) = true2.
% 5.63/1.10 Proof:
% 5.63/1.10 neq(x, nil)
% 5.63/1.10 = { by axiom 1 (co1_3) R->L }
% 5.63/1.10 neq(v, nil)
% 5.63/1.10 = { by axiom 3 (co1_5) }
% 5.63/1.10 true2
% 5.63/1.10
% 5.63/1.10 Lemma 19: ssItem(z) = true2.
% 5.63/1.10 Proof:
% 5.63/1.10 ssItem(z)
% 5.63/1.10 = { by axiom 14 (co1_15) R->L }
% 5.63/1.10 fresh16(neq(x, nil), true2)
% 5.63/1.10 = { by lemma 18 }
% 5.63/1.10 fresh16(true2, true2)
% 5.63/1.10 = { by axiom 6 (co1_15) }
% 5.63/1.10 true2
% 5.63/1.10
% 5.63/1.10 Lemma 20: ssList(x1) = true2.
% 5.63/1.10 Proof:
% 5.63/1.10 ssList(x1)
% 5.63/1.10 = { by axiom 15 (co1_17) R->L }
% 5.63/1.10 fresh14(neq(x, nil), true2)
% 5.63/1.10 = { by lemma 18 }
% 5.63/1.10 fresh14(true2, true2)
% 5.63/1.10 = { by axiom 7 (co1_17) }
% 5.63/1.10 true2
% 5.63/1.10
% 5.63/1.10 Goal 1 (co1_11): tuple5(cons(X, nil), ssItem(X), neq(x, nil), memberP(v, X)) = tuple5(u, true2, true2, true2).
% 5.63/1.10 The goal is true when:
% 5.63/1.10 X = z
% 5.63/1.10
% 5.63/1.10 Proof:
% 5.63/1.10 tuple5(cons(z, nil), ssItem(z), neq(x, nil), memberP(v, z))
% 5.63/1.10 = { by lemma 18 }
% 5.63/1.10 tuple5(cons(z, nil), ssItem(z), true2, memberP(v, z))
% 5.63/1.11 = { by axiom 1 (co1_3) }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, memberP(x, z))
% 5.63/1.11 = { by axiom 4 (co1_13) R->L }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, memberP(fresh18(true2, true2), z))
% 5.63/1.11 = { by lemma 18 R->L }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, memberP(fresh18(neq(x, nil), true2), z))
% 5.63/1.11 = { by axiom 12 (co1_13) }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, memberP(app(cons(z, nil), x1), z))
% 5.63/1.11 = { by axiom 10 (ax81) R->L }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, memberP(fresh30(true2, true2, x1, z), z))
% 5.63/1.11 = { by lemma 20 R->L }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, memberP(fresh30(ssList(x1), true2, x1, z), z))
% 5.63/1.11 = { by axiom 17 (ax81) }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, memberP(fresh29(ssItem(z), true2, x1, z), z))
% 5.63/1.11 = { by lemma 19 }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, memberP(fresh29(true2, true2, x1, z), z))
% 5.63/1.11 = { by axiom 11 (ax81) }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, memberP(cons(z, x1), z))
% 5.63/1.11 = { by axiom 8 (ax37_2) R->L }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, fresh69(true2, true2, z, x1))
% 5.63/1.11 = { by lemma 20 R->L }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, fresh69(ssList(x1), true2, z, x1))
% 5.63/1.11 = { by axiom 16 (ax37_2) }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, fresh68(ssItem(z), true2, z, x1))
% 5.63/1.11 = { by lemma 19 }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, fresh68(true2, true2, z, x1))
% 5.63/1.11 = { by axiom 9 (ax37_2) }
% 5.63/1.11 tuple5(cons(z, nil), ssItem(z), true2, true2)
% 5.63/1.11 = { by axiom 13 (co1_14) R->L }
% 5.63/1.11 tuple5(fresh17(neq(x, nil), true2), ssItem(z), true2, true2)
% 5.63/1.11 = { by lemma 18 }
% 5.63/1.11 tuple5(fresh17(true2, true2), ssItem(z), true2, true2)
% 5.63/1.11 = { by axiom 5 (co1_14) }
% 5.63/1.11 tuple5(w, ssItem(z), true2, true2)
% 5.63/1.11 = { by axiom 2 (co1_2) R->L }
% 5.63/1.11 tuple5(u, ssItem(z), true2, true2)
% 5.63/1.11 = { by lemma 19 }
% 5.63/1.11 tuple5(u, true2, true2, true2)
% 5.63/1.11 % SZS output end Proof
% 5.63/1.11
% 5.63/1.11 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------