TSTP Solution File: SWC411-1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : SWC411-1 : 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 : n001.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 20:55:27 EDT 2023

% Result   : Unsatisfiable 2.53s 0.86s
% Output   : Proof 2.53s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.14  % Problem  : SWC411-1 : TPTP v8.1.2. Released v2.4.0.
% 0.00/0.15  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.15/0.37  % Computer : n001.cluster.edu
% 0.15/0.37  % Model    : x86_64 x86_64
% 0.15/0.37  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.37  % Memory   : 8042.1875MB
% 0.15/0.37  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.15/0.37  % CPULimit : 300
% 0.15/0.37  % WCLimit  : 300
% 0.15/0.37  % DateTime : Mon Aug 28 15:17:50 EDT 2023
% 0.15/0.37  % CPUTime  : 
% 2.53/0.86  Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 2.53/0.86  
% 2.53/0.86  % SZS status Unsatisfiable
% 2.53/0.86  
% 2.53/0.86  % SZS output start Proof
% 2.53/0.86  Take the following subset of the input axioms:
% 2.53/0.86    fof(co1_10, negated_conjecture, ~memberP(sk1, sk5)).
% 2.53/0.86    fof(co1_5, negated_conjecture, sk2=sk4).
% 2.53/0.86    fof(co1_6, negated_conjecture, sk1=sk3).
% 2.53/0.86    fof(co1_7, negated_conjecture, ![A2]: (~ssItem(A2) | (memberP(sk3, A2) | ~memberP(sk4, A2)))).
% 2.53/0.86    fof(co1_8, negated_conjecture, ssItem(sk5)).
% 2.53/0.86    fof(co1_9, negated_conjecture, memberP(sk2, sk5)).
% 2.53/0.86  
% 2.53/0.86  Now clausify the problem and encode Horn clauses using encoding 3 of
% 2.53/0.86  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 2.53/0.86  We repeatedly replace C & s=t => u=v by the two clauses:
% 2.53/0.86    fresh(y, y, x1...xn) = u
% 2.53/0.86    C => fresh(s, t, x1...xn) = v
% 2.53/0.86  where fresh is a fresh function symbol and x1..xn are the free
% 2.53/0.86  variables of u and v.
% 2.53/0.86  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 2.53/0.86  input problem has no model of domain size 1).
% 2.53/0.86  
% 2.53/0.86  The encoding turns the above axioms into the following unit equations and goals:
% 2.53/0.86  
% 2.53/0.86  Axiom 1 (co1_6): sk1 = sk3.
% 2.53/0.86  Axiom 2 (co1_5): sk2 = sk4.
% 2.53/0.86  Axiom 3 (co1_8): ssItem(sk5) = true2.
% 2.53/0.86  Axiom 4 (co1_9): memberP(sk2, sk5) = true2.
% 2.53/0.86  Axiom 5 (co1_7): fresh15(X, X, Y) = memberP(sk3, Y).
% 2.53/0.86  Axiom 6 (co1_7): fresh14(X, X, Y) = true2.
% 2.53/0.86  Axiom 7 (co1_7): fresh15(memberP(sk4, X), true2, X) = fresh14(ssItem(X), true2, X).
% 2.53/0.86  
% 2.53/0.86  Goal 1 (co1_10): memberP(sk1, sk5) = true2.
% 2.53/0.86  Proof:
% 2.53/0.86    memberP(sk1, sk5)
% 2.53/0.86  = { by axiom 1 (co1_6) }
% 2.53/0.86    memberP(sk3, sk5)
% 2.53/0.86  = { by axiom 5 (co1_7) R->L }
% 2.53/0.86    fresh15(true2, true2, sk5)
% 2.53/0.86  = { by axiom 4 (co1_9) R->L }
% 2.53/0.86    fresh15(memberP(sk2, sk5), true2, sk5)
% 2.53/0.87  = { by axiom 2 (co1_5) }
% 2.53/0.87    fresh15(memberP(sk4, sk5), true2, sk5)
% 2.53/0.87  = { by axiom 7 (co1_7) }
% 2.53/0.87    fresh14(ssItem(sk5), true2, sk5)
% 2.53/0.87  = { by axiom 3 (co1_8) }
% 2.53/0.87    fresh14(true2, true2, sk5)
% 2.53/0.87  = { by axiom 6 (co1_7) }
% 2.53/0.87    true2
% 2.53/0.87  % SZS output end Proof
% 2.53/0.87  
% 2.53/0.87  RESULT: Unsatisfiable (the axioms are contradictory).
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