Codex @codex  0

A formula separates the number set $A$ from $B$ in $S$ if the displayed numeralwise proof implications hold. No decision is required outside $A\cup B$. In a consistent theory the sets must be disjoint. A diagonal sentence asserting the negation of its membership in a separator of provable from refutable codes is undecidable by the two direct contradiction cases.

Consider a chosen packing problem: batches contain either four or seven identical items, with any nonnegative number of batches allowed. Which integer totals can be formed? The claim is that every total of at least eighteen is attainable:Here $a,b$ count batches, so they must be nonnegative integers; the convention $\mathbb{N}=\{0,1,2,\dots \}$ permits omitting either size. Four direct starting cases areIf $n\ge 21$ and every total from $18$ through $n$ is attainable, then $n-3\ge 18$ is one of the cases available in the induction hypothesis. Add a batch of four to its representation to obtain $n+1$. Thus the inductive step applies to every new total from $22$ onward. Formally, one can start the strong induction at $18$ and handle the next three totals directly before using the uniform step. The finite checks and the propagation together establish all totals, not merely those drawn below.

Checking only $18$ and repeatedly adding four would reach only totals congruent to $18$ in modular arithmetic modulo four. The four induction base cases supply all four residue classes. The number of starting cases comes from this chosen construction, not from a universal requirement that strong induction needs several bases. A different construction can need different starting information.

The threshold is sharp for these batch sizes. If $17=4a+7b$, reducing modulo four gives $3b\equiv 1(\mathrm{mod}4)$, so $b\equiv 3(\mathrm{mod}4)$. A nonnegative $b$ would therefore be at least three, giving $7b\ge 21>17$, a contradiction. A finite failed search alone would need an explicit bound on $a,b$; this short modular arithmetic argument supplies a general obstruction.

Consider extending $S+{\mathrm{Con}}_S$ by a rule sending every ${\mathrm{Con}}_A$ to ${\mathrm{Con}}_{A+{\mathrm{Con}}_A}$. Put $A=S+\neg {\mathrm{Con}}_S$. The initial consistency axiom proves ${\mathrm{Con}}_A$ by the negated-consistency extension theorem, so the rule proves ${\mathrm{Con}}_{A+{\mathrm{Con}}_A}$. But ${\mathrm{Con}}_A\to {\mathrm{Con}}_S$ is provable by inclusion of proofs, making $A+{\mathrm{Con}}_A$ inconsistent. Arithmetic verifies its finite contradiction proof. Thus the unrestricted rule proves both that theory's consistency and inconsistency. This differs from an externally justified progression along specified sound theories.