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    Carmelics

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    Home/Original/inverse
    See Original
    Inverse View

    It is not the case that 3-SAT can be reduced to INDEPENDENT SET in polynomial time

    ?Set your confidence on the premises below to see your aggregate.

    Reasons For

    2 perspectives
    Reason for 1 of 2
    ?
    • 1.The reduction assumes a classical, bivalent semantics where every literal is either true or false, excluding paraconsistent or many-valued logical frameworks.
      ?

      Think about whether this reason is strong or weak

    • 2.In a many-valued logic (e.g., Łukasiewicz's three-valued system), a literal can take an intermediate value, so 'at least one literal true per clause' fails to uniquely determine a valid independent set node.
      ?

      Think about whether this reason is strong or weak

    • 3.Therefore, the bijection between satisfying valuations and independent sets of size n breaks down in any non-bivalent semantic framework, limiting the reduction's logical generality.
      ?

      Think about whether this reason is strong or weak

    Reason for 2 of 2
    ?
    • 1.The argument conflates syntactic polynomial-time constructibility of G_φ with a semantic guarantee that the graph faithfully encodes all and only satisfying assignments.
      ?

      Think about whether this reason is strong or weak

    • 2.Quine's thesis that the boundary between logical truth and empirical fact is indeterminate implies that the 'edges between contradictory literals' encode a notion of contradiction that is framework-relative, not absolute.
      ?

      Think about whether this reason is strong or weak

    • 3.If the identification of contradictory literal pairs presupposes a fixed, non-negotiable logical syntax, the reduction embeds a substantive metaphysical assumption about negation that is not itself established by the polynomial-time construction.
      ?

      Think about whether this reason is strong or weak

    Reasons Against

    1 perspective
    Reason against
    ?
    • 1.A graph G_φ with n triangles can be constructed from a 3-CNF formula φ with n clauses in polynomial time (O(n²) edges from 3n vertices)
      ?

      Think about whether this reason is strong or weak

    • 2.If φ is satisfiable, then a satisfying valuation makes at least one literal true per clause, yielding an independent set of size n in G_φ by selecting one node per triangle
      ?

      Think about whether this reason is strong or weak

    • 3.If G_φ has an independent set of size n, then by construction it contains exactly one node per triangle, and edges between contradictory literals across triangles prevent contradictions, so a satisfying valuation for φ can be constructed
      ?

      Think about whether this reason is strong or weak

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    Strongest counterpoint
    Explore the most compelling reason on the other side.