TSTP Solution File: SWC406+1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : SWC406+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 20:55:25 EDT 2023

% Result   : Theorem 3.18s 0.86s
% Output   : Proof 3.98s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.12  % Problem  : SWC406+1 : TPTP v8.1.2. Released v2.4.0.
% 0.12/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35  % Computer : n016.cluster.edu
% 0.13/0.35  % Model    : x86_64 x86_64
% 0.13/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35  % Memory   : 8042.1875MB
% 0.13/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35  % CPULimit : 300
% 0.13/0.35  % WCLimit  : 300
% 0.13/0.35  % DateTime : Mon Aug 28 18:54:29 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 3.18/0.86  Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 3.18/0.86  
% 3.18/0.86  % SZS status Theorem
% 3.18/0.86  
% 3.18/0.86  % SZS output start Proof
% 3.18/0.86  Take the following subset of the input axioms:
% 3.98/0.86    fof(co1, conjecture, ![U]: (ssList(U) => ![V]: (ssList(V) => ![W]: (ssList(W) => ![X]: (ssList(X) => (V!=X | (U!=W | (?[Y]: (ssItem(Y) & ((~memberP(W, Y) & (![Z]: (ssItem(Z) => (~memberP(X, Z) | (~leq(Z, Y) | Y=Z))) & memberP(X, Y))) | (memberP(W, Y) & (~memberP(X, Y) | ?[Z2]: (ssItem(Z2) & (Y!=Z2 & (memberP(X, Z2) & leq(Z2, Y)))))))) | ![X1]: (ssItem(X1) => (~memberP(U, X1) | memberP(V, X1))))))))))).
% 3.98/0.86  
% 3.98/0.86  Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.98/0.86  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.98/0.86  We repeatedly replace C & s=t => u=v by the two clauses:
% 3.98/0.86    fresh(y, y, x1...xn) = u
% 3.98/0.86    C => fresh(s, t, x1...xn) = v
% 3.98/0.86  where fresh is a fresh function symbol and x1..xn are the free
% 3.98/0.86  variables of u and v.
% 3.98/0.86  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.98/0.86  input problem has no model of domain size 1).
% 3.98/0.86  
% 3.98/0.86  The encoding turns the above axioms into the following unit equations and goals:
% 3.98/0.86  
% 3.98/0.86  Axiom 1 (co1_1): v = x.
% 3.98/0.86  Axiom 2 (co1): u = w.
% 3.98/0.86  Axiom 3 (co1_2): ssItem(x1) = true2.
% 3.98/0.86  Axiom 4 (co1_7): memberP(u, x1) = true2.
% 3.98/0.86  Axiom 5 (co1_10): fresh16(X, X, Y) = memberP(x, Y).
% 3.98/0.86  Axiom 6 (co1_10): fresh15(X, X, Y) = true2.
% 3.98/0.86  Axiom 7 (co1_10): fresh16(memberP(w, X), true2, X) = fresh15(ssItem(X), true2, X).
% 3.98/0.86  
% 3.98/0.86  Goal 1 (co1_14): memberP(v, x1) = true2.
% 3.98/0.86  Proof:
% 3.98/0.86    memberP(v, x1)
% 3.98/0.86  = { by axiom 1 (co1_1) }
% 3.98/0.86    memberP(x, x1)
% 3.98/0.86  = { by axiom 5 (co1_10) R->L }
% 3.98/0.86    fresh16(true2, true2, x1)
% 3.98/0.86  = { by axiom 4 (co1_7) R->L }
% 3.98/0.86    fresh16(memberP(u, x1), true2, x1)
% 3.98/0.86  = { by axiom 2 (co1) }
% 3.98/0.86    fresh16(memberP(w, x1), true2, x1)
% 3.98/0.86  = { by axiom 7 (co1_10) }
% 3.98/0.86    fresh15(ssItem(x1), true2, x1)
% 3.98/0.86  = { by axiom 3 (co1_2) }
% 3.98/0.86    fresh15(true2, true2, x1)
% 3.98/0.86  = { by axiom 6 (co1_10) }
% 3.98/0.86    true2
% 3.98/0.86  % SZS output end Proof
% 3.98/0.86  
% 3.98/0.86  RESULT: Theorem (the conjecture is true).
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