TSTP Solution File: SWC102-1 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWC102-1 : 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 : n008.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 20:53:52 EDT 2023

% Result   : Unsatisfiable 3.51s 0.86s
% Output   : Proof 3.51s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : SWC102-1 : 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 : n008.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 14:42:32 EDT 2023
% 0.12/0.34  % CPUTime  : 
% 3.51/0.86  Command-line arguments: --flatten
% 3.51/0.86  
% 3.51/0.86  % SZS status Unsatisfiable
% 3.51/0.86  
% 3.51/0.87  % SZS output start Proof
% 3.51/0.87  Take the following subset of the input axioms:
% 3.51/0.87    fof(clause1, axiom, equalelemsP(nil)).
% 3.51/0.87    fof(clause2, axiom, duplicatefreeP(nil)).
% 3.51/0.87    fof(clause61, axiom, ![U]: (~ssList(U) | frontsegP(U, U))).
% 3.51/0.87    fof(co1_3, negated_conjecture, ssList(sk3)).
% 3.51/0.87    fof(co1_5, negated_conjecture, sk2=sk4).
% 3.51/0.87    fof(co1_6, negated_conjecture, sk1=sk3).
% 3.51/0.87    fof(co1_7, negated_conjecture, sk4=sk3).
% 3.51/0.87    fof(co1_8, negated_conjecture, neq(sk2, nil)).
% 3.51/0.87    fof(co1_9, negated_conjecture, ~neq(sk1, nil) | ~frontsegP(sk2, sk1)).
% 3.51/0.87  
% 3.51/0.87  Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.51/0.87  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.51/0.87  We repeatedly replace C & s=t => u=v by the two clauses:
% 3.51/0.87    fresh(y, y, x1...xn) = u
% 3.51/0.87    C => fresh(s, t, x1...xn) = v
% 3.51/0.87  where fresh is a fresh function symbol and x1..xn are the free
% 3.51/0.87  variables of u and v.
% 3.51/0.87  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.51/0.87  input problem has no model of domain size 1).
% 3.51/0.87  
% 3.51/0.87  The encoding turns the above axioms into the following unit equations and goals:
% 3.51/0.87  
% 3.51/0.87  Axiom 1 (co1_7): sk4 = sk3.
% 3.51/0.87  Axiom 2 (co1_5): sk2 = sk4.
% 3.51/0.87  Axiom 3 (co1_6): sk1 = sk3.
% 3.51/0.87  Axiom 4 (clause2): duplicatefreeP(nil) = true2.
% 3.51/0.87  Axiom 5 (clause1): equalelemsP(nil) = true2.
% 3.51/0.87  Axiom 6 (co1_3): ssList(sk3) = true2.
% 3.51/0.87  Axiom 7 (co1_8): neq(sk2, nil) = true2.
% 3.51/0.87  Axiom 8 (clause61): fresh51(X, X, Y) = true2.
% 3.51/0.87  Axiom 9 (clause61): fresh51(ssList(X), true2, X) = frontsegP(X, X).
% 3.51/0.87  
% 3.51/0.87  Lemma 10: sk2 = sk3.
% 3.51/0.87  Proof:
% 3.51/0.87    sk2
% 3.51/0.87  = { by axiom 2 (co1_5) }
% 3.51/0.87    sk4
% 3.51/0.87  = { by axiom 1 (co1_7) }
% 3.51/0.87    sk3
% 3.51/0.87  
% 3.51/0.87  Lemma 11: equalelemsP(nil) = duplicatefreeP(nil).
% 3.51/0.87  Proof:
% 3.51/0.87    equalelemsP(nil)
% 3.51/0.87  = { by axiom 5 (clause1) }
% 3.51/0.87    true2
% 3.51/0.87  = { by axiom 4 (clause2) R->L }
% 3.51/0.87    duplicatefreeP(nil)
% 3.51/0.87  
% 3.51/0.87  Lemma 12: frontsegP(sk2, sk1) = frontsegP(sk3, sk3).
% 3.51/0.87  Proof:
% 3.51/0.87    frontsegP(sk2, sk1)
% 3.51/0.87  = { by lemma 10 }
% 3.51/0.87    frontsegP(sk3, sk1)
% 3.51/0.87  = { by axiom 3 (co1_6) }
% 3.51/0.87    frontsegP(sk3, sk3)
% 3.51/0.87  
% 3.51/0.87  Goal 1 (co1_9): tuple2(frontsegP(sk2, sk1), neq(sk1, nil)) = tuple2(true2, true2).
% 3.51/0.87  Proof:
% 3.51/0.87    tuple2(frontsegP(sk2, sk1), neq(sk1, nil))
% 3.51/0.87  = { by lemma 10 }
% 3.51/0.87    tuple2(frontsegP(sk3, sk1), neq(sk1, nil))
% 3.51/0.87  = { by axiom 3 (co1_6) }
% 3.51/0.87    tuple2(frontsegP(sk3, sk1), neq(sk3, nil))
% 3.51/0.87  = { by axiom 3 (co1_6) }
% 3.51/0.87    tuple2(frontsegP(sk3, sk3), neq(sk3, nil))
% 3.51/0.87  = { by lemma 12 R->L }
% 3.51/0.87    tuple2(frontsegP(sk2, sk1), neq(sk3, nil))
% 3.51/0.87  = { by lemma 10 R->L }
% 3.51/0.87    tuple2(frontsegP(sk2, sk1), neq(sk2, nil))
% 3.51/0.87  = { by axiom 7 (co1_8) }
% 3.51/0.87    tuple2(frontsegP(sk2, sk1), true2)
% 3.51/0.87  = { by axiom 4 (clause2) R->L }
% 3.51/0.87    tuple2(frontsegP(sk2, sk1), duplicatefreeP(nil))
% 3.51/0.87  = { by lemma 11 R->L }
% 3.51/0.87    tuple2(frontsegP(sk2, sk1), equalelemsP(nil))
% 3.51/0.87  = { by lemma 12 }
% 3.51/0.87    tuple2(frontsegP(sk3, sk3), equalelemsP(nil))
% 3.51/0.87  = { by axiom 9 (clause61) R->L }
% 3.51/0.87    tuple2(fresh51(ssList(sk3), true2, sk3), equalelemsP(nil))
% 3.51/0.87  = { by axiom 4 (clause2) R->L }
% 3.51/0.87    tuple2(fresh51(ssList(sk3), duplicatefreeP(nil), sk3), equalelemsP(nil))
% 3.51/0.87  = { by lemma 11 R->L }
% 3.51/0.87    tuple2(fresh51(ssList(sk3), equalelemsP(nil), sk3), equalelemsP(nil))
% 3.51/0.87  = { by axiom 6 (co1_3) }
% 3.51/0.87    tuple2(fresh51(true2, equalelemsP(nil), sk3), equalelemsP(nil))
% 3.51/0.87  = { by axiom 4 (clause2) R->L }
% 3.51/0.87    tuple2(fresh51(duplicatefreeP(nil), equalelemsP(nil), sk3), equalelemsP(nil))
% 3.51/0.87  = { by lemma 11 R->L }
% 3.51/0.87    tuple2(fresh51(equalelemsP(nil), equalelemsP(nil), sk3), equalelemsP(nil))
% 3.51/0.87  = { by axiom 8 (clause61) }
% 3.51/0.87    tuple2(true2, equalelemsP(nil))
% 3.51/0.87  = { by axiom 4 (clause2) R->L }
% 3.51/0.88    tuple2(duplicatefreeP(nil), equalelemsP(nil))
% 3.51/0.88  = { by lemma 11 }
% 3.51/0.88    tuple2(duplicatefreeP(nil), duplicatefreeP(nil))
% 3.51/0.88  = { by axiom 4 (clause2) }
% 3.51/0.88    tuple2(true2, duplicatefreeP(nil))
% 3.51/0.88  = { by axiom 4 (clause2) }
% 3.51/0.88    tuple2(true2, true2)
% 3.51/0.88  % SZS output end Proof
% 3.51/0.88  
% 3.51/0.88  RESULT: Unsatisfiable (the axioms are contradictory).
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