TSTP Solution File: SWC007+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWC007+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 : n016.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 20:53:22 EDT 2023
% Result : Theorem 5.36s 1.05s
% Output : Proof 5.36s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.12 % Problem : SWC007+1 : TPTP v8.1.2. Released v2.4.0.
% 0.04/0.13 % 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 : n016.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 15:33:29 EDT 2023
% 0.12/0.34 % CPUTime :
% 5.36/1.05 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 5.36/1.05
% 5.36/1.05 % SZS status Theorem
% 5.36/1.05
% 5.36/1.05 % SZS output start Proof
% 5.36/1.05 Take the following subset of the input axioms:
% 5.36/1.06 fof(ax1, axiom, ![U]: (ssItem(U) => ![V]: (ssItem(V) => (neq(U, V) <=> U!=V)))).
% 5.36/1.06 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))))))))).
% 5.36/1.06 fof(ax15, axiom, ![U2]: (ssList(U2) => ![V2]: (ssList(V2) => (neq(U2, V2) <=> U2!=V2)))).
% 5.36/1.06 fof(ax17, axiom, ssList(nil)).
% 5.36/1.06 fof(ax18, axiom, ![U2]: (ssList(U2) => ![V2]: (ssItem(V2) => cons(V2, U2)!=U2))).
% 5.36/1.06 fof(ax21, axiom, ![U2]: (ssList(U2) => ![V2]: (ssItem(V2) => nil!=cons(V2, U2)))).
% 5.36/1.06 fof(ax28, axiom, ![U2]: (ssList(U2) => app(nil, U2)=U2)).
% 5.36/1.06 fof(ax33, axiom, ![U2]: (ssItem(U2) => ![V2]: (ssItem(V2) => (lt(U2, V2) => ~lt(V2, U2))))).
% 5.36/1.06 fof(ax38, axiom, ![U2]: (ssItem(U2) => ~memberP(nil, U2))).
% 5.36/1.06 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))))))))))).
% 5.36/1.06 fof(ax84, axiom, ![U2]: (ssList(U2) => app(U2, nil)=U2)).
% 5.36/1.06 fof(ax90, axiom, ![U2]: (ssItem(U2) => ~lt(U2, U2))).
% 5.36/1.06 fof(ax93, axiom, ![U2]: (ssItem(U2) => ![V2]: (ssItem(V2) => (lt(U2, V2) <=> (U2!=V2 & leq(U2, V2)))))).
% 5.36/1.06 fof(ax94, axiom, ![U2]: (ssItem(U2) => ![V2]: (ssItem(V2) => (gt(U2, V2) => ~gt(V2, U2))))).
% 5.36/1.06 fof(co1, conjecture, ![U2]: (ssList(U2) => ![V2]: (ssList(V2) => ![W2]: (ssList(W2) => ![X2]: (ssList(X2) => (nil!=W2 | (V2!=X2 | (U2!=W2 | ?[Y2]: (ssList(Y2) & ?[Z2]: (ssList(Z2) & ?[X1]: (ssList(X1) & (app(app(Y2, Z2), X1)=V2 & app(Y2, X1)=U2)))))))))))).
% 5.36/1.06
% 5.36/1.06 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.36/1.06 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.36/1.06 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.36/1.06 fresh(y, y, x1...xn) = u
% 5.36/1.06 C => fresh(s, t, x1...xn) = v
% 5.36/1.06 where fresh is a fresh function symbol and x1..xn are the free
% 5.36/1.06 variables of u and v.
% 5.36/1.06 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.36/1.06 input problem has no model of domain size 1).
% 5.36/1.06
% 5.36/1.06 The encoding turns the above axioms into the following unit equations and goals:
% 5.36/1.06
% 5.36/1.06 Axiom 1 (co1): nil = w.
% 5.36/1.06 Axiom 2 (co1_1): u = w.
% 5.36/1.06 Axiom 3 (ax17): ssList(nil) = true2.
% 5.36/1.06 Axiom 4 (co1_4): ssList(v) = true2.
% 5.36/1.06 Axiom 5 (ax84): fresh(X, X, Y) = Y.
% 5.36/1.06 Axiom 6 (ax28): fresh7(X, X, Y) = Y.
% 5.36/1.06 Axiom 7 (ax84): fresh(ssList(X), true2, X) = app(X, nil).
% 5.36/1.06 Axiom 8 (ax28): fresh7(ssList(X), true2, X) = app(nil, X).
% 5.36/1.06
% 5.36/1.06 Goal 1 (co1_7): tuple6(app(X, Y), app(app(X, Z), Y), ssList(X), ssList(Z), ssList(Y)) = tuple6(u, v, true2, true2, true2).
% 5.36/1.06 The goal is true when:
% 5.36/1.06 X = nil
% 5.36/1.06 Y = nil
% 5.36/1.06 Z = v
% 5.36/1.06
% 5.36/1.06 Proof:
% 5.36/1.06 tuple6(app(nil, nil), app(app(nil, v), nil), ssList(nil), ssList(v), ssList(nil))
% 5.36/1.06 = { by axiom 4 (co1_4) }
% 5.36/1.06 tuple6(app(nil, nil), app(app(nil, v), nil), ssList(nil), true2, ssList(nil))
% 5.36/1.06 = { by axiom 7 (ax84) R->L }
% 5.36/1.06 tuple6(fresh(ssList(nil), true2, nil), app(app(nil, v), nil), ssList(nil), true2, ssList(nil))
% 5.36/1.06 = { by axiom 3 (ax17) }
% 5.36/1.06 tuple6(fresh(true2, true2, nil), app(app(nil, v), nil), ssList(nil), true2, ssList(nil))
% 5.36/1.06 = { by axiom 5 (ax84) }
% 5.36/1.06 tuple6(nil, app(app(nil, v), nil), ssList(nil), true2, ssList(nil))
% 5.36/1.06 = { by axiom 8 (ax28) R->L }
% 5.36/1.06 tuple6(nil, app(fresh7(ssList(v), true2, v), nil), ssList(nil), true2, ssList(nil))
% 5.36/1.06 = { by axiom 4 (co1_4) }
% 5.36/1.06 tuple6(nil, app(fresh7(true2, true2, v), nil), ssList(nil), true2, ssList(nil))
% 5.36/1.06 = { by axiom 6 (ax28) }
% 5.36/1.06 tuple6(nil, app(v, nil), ssList(nil), true2, ssList(nil))
% 5.36/1.06 = { by axiom 7 (ax84) R->L }
% 5.36/1.06 tuple6(nil, fresh(ssList(v), true2, v), ssList(nil), true2, ssList(nil))
% 5.36/1.06 = { by axiom 4 (co1_4) }
% 5.36/1.06 tuple6(nil, fresh(true2, true2, v), ssList(nil), true2, ssList(nil))
% 5.36/1.06 = { by axiom 5 (ax84) }
% 5.36/1.06 tuple6(nil, v, ssList(nil), true2, ssList(nil))
% 5.36/1.06 = { by axiom 3 (ax17) }
% 5.36/1.06 tuple6(nil, v, ssList(nil), true2, true2)
% 5.36/1.06 = { by axiom 3 (ax17) }
% 5.36/1.06 tuple6(nil, v, true2, true2, true2)
% 5.36/1.06 = { by axiom 1 (co1) }
% 5.36/1.06 tuple6(w, v, true2, true2, true2)
% 5.36/1.06 = { by axiom 2 (co1_1) R->L }
% 5.36/1.06 tuple6(u, v, true2, true2, true2)
% 5.36/1.06 % SZS output end Proof
% 5.36/1.06
% 5.36/1.06 RESULT: Theorem (the conjecture is true).
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