 Solution

GL has all classical propositional tautologies, the normal distribution axiom $\square (A\to B)\to (\square A\to \square B)$ and the Löb axiom $\square (\square A\to A)\to \square A$, closed under uniform substitution, modus ponens and necessitation of theorems. We use its derived positive-introspection theorem $\square A\to \square \square A$, normal boxed reasoning, and the boxed replacement principle supplied in the question.

For completeness, positive introspection follows from these axioms. Put $C=A\wedge \square A$. Distribution gives $\square C\to \square A$, so propositional logic proves $A\to (\square C\to C)$. Necessitate and distribute to get $\square A\to \square (\square C\to C)$; Löb's axiom gives $\square A\to \square C$, and boxing $C\to \square A$ finishes $\square A\to \square \square A$.

For a GL fixed point by truth substitution, set $\top =(\perp \to \perp )$, $U=\square Y(\top )$ and $V=\square Y(U)$. From the tautology $U\to (\top \leftrightarrow U)$, boxed reasoning gives $\square U\to \square (\top \leftrightarrow U)$. The supplied replacement principle and distribution then giveSince $U$ is itself boxed, positive introspection gives $U\to \square U$, and hence $U\to V$.

For the reverse direction, the displayed implication entails the theorem $V\to (\square U\to U)$. Necessitate it and distribute to obtain $\square V\to \square (\square U\to U)$. Since $V$ is boxed, $V\to \square V$ holds. The Löb axiom therefore yields $V\to \square U$, and the same displayed implication then yields $V\to U$. ConsequentlyAll occurrences of the unknown letter have been removed from $U$ by substitution of truth; any other parameter letters remain available. The proof does not assume the invalid reflection principle $\square U\to U$ as an unconditional axiom.