TSTP Solution File: SWC012+1 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWC012+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 : n007.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 20:53:25 EDT 2023

% Result   : Theorem 6.70s 1.20s
% Output   : Proof 6.70s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.11  % Problem  : SWC012+1 : TPTP v8.1.2. Released v2.4.0.
% 0.03/0.12  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.33  % Computer : n007.cluster.edu
% 0.12/0.33  % Model    : x86_64 x86_64
% 0.12/0.33  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33  % Memory   : 8042.1875MB
% 0.12/0.33  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33  % CPULimit : 300
% 0.12/0.33  % WCLimit  : 300
% 0.12/0.33  % DateTime : Mon Aug 28 16:07:13 EDT 2023
% 0.12/0.33  % CPUTime  : 
% 6.70/1.20  Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 6.70/1.20  
% 6.70/1.20  % SZS status Theorem
% 6.70/1.20  
% 6.70/1.20  % SZS output start Proof
% 6.70/1.20  Take the following subset of the input axioms:
% 6.70/1.21    fof(ax1, axiom, ![U]: (ssItem(U) => ![V]: (ssItem(V) => (neq(U, V) <=> U!=V)))).
% 6.70/1.21    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))))))))).
% 6.70/1.21    fof(ax15, axiom, ![U2]: (ssList(U2) => ![V2]: (ssList(V2) => (neq(U2, V2) <=> U2!=V2)))).
% 6.70/1.21    fof(ax17, axiom, ssList(nil)).
% 6.70/1.21    fof(ax18, axiom, ![U2]: (ssList(U2) => ![V2]: (ssItem(V2) => cons(V2, U2)!=U2))).
% 6.70/1.21    fof(ax21, axiom, ![U2]: (ssList(U2) => ![V2]: (ssItem(V2) => nil!=cons(V2, U2)))).
% 6.70/1.21    fof(ax33, axiom, ![U2]: (ssItem(U2) => ![V2]: (ssItem(V2) => (lt(U2, V2) => ~lt(V2, U2))))).
% 6.70/1.21    fof(ax38, axiom, ![U2]: (ssItem(U2) => ~memberP(nil, U2))).
% 6.70/1.21    fof(ax8, axiom, ![U2]: (ssList(U2) => (cyclefreeP(U2) <=> ![V2]: (ssItem(V2) => ![W2]: (ssItem(W2) => ![X3]: (ssList(X3) => ![Y2]: (ssList(Y2) => ![Z2]: (ssList(Z2) => (app(app(X3, cons(V2, Y2)), cons(W2, Z2))=U2 => ~(leq(V2, W2) & leq(W2, V2))))))))))).
% 6.70/1.21    fof(ax84, axiom, ![U2]: (ssList(U2) => app(U2, nil)=U2)).
% 6.70/1.21    fof(ax90, axiom, ![U2]: (ssItem(U2) => ~lt(U2, U2))).
% 6.70/1.21    fof(ax93, axiom, ![U2]: (ssItem(U2) => ![V2]: (ssItem(V2) => (lt(U2, V2) <=> (U2!=V2 & leq(U2, V2)))))).
% 6.70/1.21    fof(ax94, axiom, ![U2]: (ssItem(U2) => ![V2]: (ssItem(V2) => (gt(U2, V2) => ~gt(V2, U2))))).
% 6.70/1.21    fof(co1, conjecture, ![U2]: (ssList(U2) => ![V2]: (ssList(V2) => ![W2]: (ssList(W2) => ![X3]: (ssList(X3) => (V2!=X3 | (U2!=W2 | ((~neq(V2, nil) | (?[Y2]: (ssItem(Y2) & ?[Z2]: (ssList(Z2) & ?[X1]: (ssList(X1) & (app(app(Z2, cons(Y2, nil)), X1)=V2 & app(Z2, X1)=U2)))) | ![X2]: (ssItem(X2) => app(W2, cons(X2, nil))!=X3))) & (~neq(V2, nil) | neq(X3, nil)))))))))).
% 6.70/1.21  
% 6.70/1.21  Now clausify the problem and encode Horn clauses using encoding 3 of
% 6.70/1.21  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 6.70/1.21  We repeatedly replace C & s=t => u=v by the two clauses:
% 6.70/1.21    fresh(y, y, x1...xn) = u
% 6.70/1.21    C => fresh(s, t, x1...xn) = v
% 6.70/1.21  where fresh is a fresh function symbol and x1..xn are the free
% 6.70/1.21  variables of u and v.
% 6.70/1.21  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 6.70/1.21  input problem has no model of domain size 1).
% 6.70/1.21  
% 6.70/1.21  The encoding turns the above axioms into the following unit equations and goals:
% 6.70/1.21  
% 6.70/1.21  Axiom 1 (co1_2): v = x.
% 6.70/1.21  Axiom 2 (co1_1): u = w.
% 6.70/1.21  Axiom 3 (ax17): ssList(nil) = true2.
% 6.70/1.21  Axiom 4 (co1_6): ssList(v) = true2.
% 6.70/1.21  Axiom 5 (co1_5): ssList(u) = true2.
% 6.70/1.21  Axiom 6 (co1_4): neq(v, nil) = true2.
% 6.70/1.21  Axiom 7 (co1_11): fresh16(X, X) = x.
% 6.70/1.21  Axiom 8 (co1_12): fresh15(X, X) = true2.
% 6.70/1.21  Axiom 9 (ax84): fresh(X, X, Y) = Y.
% 6.70/1.21  Axiom 10 (ax84): fresh(ssList(X), true2, X) = app(X, nil).
% 6.70/1.21  Axiom 11 (co1_11): fresh16(neq(x, nil), true2) = app(w, cons(x2, nil)).
% 6.70/1.21  Axiom 12 (co1_12): fresh15(neq(x, nil), true2) = ssItem(x2).
% 6.70/1.21  
% 6.70/1.21  Goal 1 (co1_9): tuple6(app(X, Y), app(app(X, cons(Z, nil)), Y), ssItem(Z), neq(x, nil), ssList(X), ssList(Y)) = tuple6(u, v, true2, true2, true2, true2).
% 6.70/1.21  The goal is true when:
% 6.70/1.21    X = u
% 6.70/1.21    Y = nil
% 6.70/1.21    Z = x2
% 6.70/1.21  
% 6.70/1.21  Proof:
% 6.70/1.21    tuple6(app(u, nil), app(app(u, cons(x2, nil)), nil), ssItem(x2), neq(x, nil), ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 1 (co1_2) R->L }
% 6.70/1.21    tuple6(app(u, nil), app(app(u, cons(x2, nil)), nil), ssItem(x2), neq(v, nil), ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 6 (co1_4) }
% 6.70/1.21    tuple6(app(u, nil), app(app(u, cons(x2, nil)), nil), ssItem(x2), true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 2 (co1_1) }
% 6.70/1.21    tuple6(app(u, nil), app(app(w, cons(x2, nil)), nil), ssItem(x2), true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 11 (co1_11) R->L }
% 6.70/1.21    tuple6(app(u, nil), app(fresh16(neq(x, nil), true2), nil), ssItem(x2), true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 1 (co1_2) R->L }
% 6.70/1.21    tuple6(app(u, nil), app(fresh16(neq(v, nil), true2), nil), ssItem(x2), true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 6 (co1_4) }
% 6.70/1.21    tuple6(app(u, nil), app(fresh16(true2, true2), nil), ssItem(x2), true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 7 (co1_11) }
% 6.70/1.21    tuple6(app(u, nil), app(x, nil), ssItem(x2), true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 1 (co1_2) R->L }
% 6.70/1.21    tuple6(app(u, nil), app(v, nil), ssItem(x2), true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 12 (co1_12) R->L }
% 6.70/1.21    tuple6(app(u, nil), app(v, nil), fresh15(neq(x, nil), true2), true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 1 (co1_2) R->L }
% 6.70/1.21    tuple6(app(u, nil), app(v, nil), fresh15(neq(v, nil), true2), true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 6 (co1_4) }
% 6.70/1.21    tuple6(app(u, nil), app(v, nil), fresh15(true2, true2), true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 8 (co1_12) }
% 6.70/1.21    tuple6(app(u, nil), app(v, nil), true2, true2, ssList(u), ssList(nil))
% 6.70/1.21  = { by axiom 5 (co1_5) }
% 6.70/1.21    tuple6(app(u, nil), app(v, nil), true2, true2, true2, ssList(nil))
% 6.70/1.21  = { by axiom 3 (ax17) }
% 6.70/1.21    tuple6(app(u, nil), app(v, nil), true2, true2, true2, true2)
% 6.70/1.21  = { by axiom 10 (ax84) R->L }
% 6.70/1.21    tuple6(fresh(ssList(u), true2, u), app(v, nil), true2, true2, true2, true2)
% 6.70/1.21  = { by axiom 5 (co1_5) }
% 6.70/1.21    tuple6(fresh(true2, true2, u), app(v, nil), true2, true2, true2, true2)
% 6.70/1.21  = { by axiom 9 (ax84) }
% 6.70/1.21    tuple6(u, app(v, nil), true2, true2, true2, true2)
% 6.70/1.21  = { by axiom 10 (ax84) R->L }
% 6.70/1.21    tuple6(u, fresh(ssList(v), true2, v), true2, true2, true2, true2)
% 6.70/1.21  = { by axiom 4 (co1_6) }
% 6.70/1.21    tuple6(u, fresh(true2, true2, v), true2, true2, true2, true2)
% 6.70/1.21  = { by axiom 9 (ax84) }
% 6.70/1.21    tuple6(u, v, true2, true2, true2, true2)
% 6.70/1.22  % SZS output end Proof
% 6.70/1.22  
% 6.70/1.22  RESULT: Theorem (the conjecture is true).
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