A sound effective theory strong enough for arithmetic cannot decide every arithmetic sentence. I want the notation, the mechanism, and the failure mode visible at the same time.
Set-up
A decision problem is computable when a Turing machine \(M_e(x)\) halts on every input with the correct answer. A set \(A\subseteq\mathbf N\) is computably enumerable when a machine can list its members.
I read the first line as input and the second as output. The symbols \(\forall\) and \(\exists\) are not interchangeable, and neither may be upgraded silently to \(\Longleftrightarrow\).
The calculation
The computation below is not a second theorem. It is a checksum for the definitions and a place to inspect the difficult LaTeX at full size.
What survives abstraction
The compact alignment is a local map of the argument: assumptions on the first row, consequence on the second. Any generalisation must preserve that dependency.
The boundary
Enumerability is weaker than decidability. A search may confirm membership eventually without ever certifying nonmembership.
The result is compact enough to reuse without pretending that the caveat has disappeared. The worked line remains the quickest consistency check.