[proofplan] We prove the cycle of implications $1\Rightarrow 3\Rightarrow 2\Rightarrow 1$. The coding clauses are used only to ensure that preservation of a vertical-section code below a large enough rank initial segment recovers the relevant values of the function. The final implication uses a least-failure function for a fixed parameter $A$; the strengthened clause 2 includes the comparison $f(\kappa)\leq j(f)(\kappa)$, so the embedding supplied by clause 2 witnesses strongness at or above the alleged least failure and gives a direct contradiction by downward restriction. [/proofplan]

[step:Apply the strongness hypothesis to the paired code] Assume clause 1. Let $A\subset V_\delta$ be a set, and let $f:\delta\to\delta$ be a function. Define $A_f:=\operatorname{Code}(f)\subset V_\delta$ and $B:=A\oplus A_f\subset V_\delta$. Clause 1 applied to $B$ gives a cardinal $\kappa<\delta$ such that, for every ordinal $\lambda$ with $\kappa<\lambda<\delta$, the cardinal $\kappa$ is $\lambda$-$B$-strong. We prove $f``\kappa\subset\kappa$. Suppose, toward a contradiction, that there is an ordinal $\alpha<\kappa$ with $f(\alpha)\geq\kappa$. Since $\delta$ is inaccessible, hence a limit ordinal, choose $\lambda<\delta$ such that $\kappa<\lambda$, $f(\alpha)<\lambda$, and $\operatorname{rank}(\langle \alpha,f(\alpha)\rangle)<\lambda$. Let $j:V\to M$ witness that $\kappa$ is $\lambda$-$B$-strong. Then $j(B)\cap V_\lambda=B\cap V_\lambda$, and the disjoint-union coding gives $j(A_f)\cap V_\lambda=A_f\cap V_\lambda$. Because $\alpha<\kappa=\operatorname{crit}(j)$, we have $j(\alpha)=\alpha$. By elementarity, \begin{align*} j(f)(\alpha)=j(f(\alpha)). \end{align*} Since $f(\alpha)\geq\kappa$, monotonicity of elementary embeddings on ordinals gives $j(f(\alpha))\geq j(\kappa)>\lambda$. Put $\beta:=f(\alpha)$. Then $\beta<\lambda$, and $\langle \alpha,\beta\rangle\in V_\lambda$. Also $\beta<j(f)(\alpha)$, so $\langle \alpha,\beta\rangle\in j(A_f)=\operatorname{Code}(j(f))$. But $\langle \alpha,\beta\rangle\notin A_f=\operatorname{Code}(f)$, because the vertical section of $A_f$ over $\alpha$ contains exactly the ordinals strictly below $f(\alpha)$. This contradicts $j(A_f)\cap V_\lambda=A_f\cap V_\lambda$. Therefore $f``\kappa\subset\kappa$. Since the same $\kappa$ is $\lambda$-$B$-strong for every $\kappa<\lambda<\delta$, and $B=A\oplus\operatorname{Code}(f)$, clause 3 follows. [/step]

[step:Recover the value of $j(f)$ at the critical point from the vertical-section code] Assume clause 3. Let $A\subset V_\delta$ be a set, and let $f:\delta\to\delta$ be a function. Put $B:=A\oplus\operatorname{Code}(f)$. Clause 3 gives a cardinal $\kappa<\delta$ such that $f``\kappa\subset\kappa$ and such that $\kappa$ is $\lambda$-$B$-strong for every ordinal $\lambda$ with $\kappa<\lambda<\delta$. Choose $\lambda<\delta$ such that $\kappa<\lambda$, $f(\kappa)<\lambda$, and $\operatorname{rank}(\langle \kappa,\beta\rangle)<\lambda$ for every ordinal $\beta\leq f(\kappa)$. This is possible because $f(\kappa)<\delta$ and $\delta$ is a limit cardinal. Let $j:V\to M$ witness that $\kappa$ is $\lambda$-$B$-strong. Then $\operatorname{crit}(j)=\kappa$, $j(\kappa)>\lambda$, $V_\lambda\subset M$, and $j(B)\cap V_\lambda=B\cap V_\lambda$. By the coordinate projections of the disjoint-union code, we have \begin{align*} j(A)\cap V_\lambda=A\cap V_\lambda. \end{align*} We also have \begin{align*} j(\operatorname{Code}(f))\cap V_\lambda=\operatorname{Code}(f)\cap V_\lambda. \end{align*} By elementarity of the fixed coding convention, $j(\operatorname{Code}(f))=\operatorname{Code}(j(f))$. We show that $j(f)(\kappa)=f(\kappa)$. For every $\beta<f(\kappa)$, the pair $\langle \kappa,\beta\rangle$ lies in $V_\lambda$ and in $\operatorname{Code}(f)$, hence it lies in $\operatorname{Code}(j(f))$. Therefore $\beta<j(f)(\kappa)$ for every $\beta<f(\kappa)$, so $f(\kappa)\leq j(f)(\kappa)$. Conversely, if $j(f)(\kappa)>f(\kappa)$, then $\langle \kappa,f(\kappa)\rangle$ lies in $V_\lambda$ and in $\operatorname{Code}(j(f))$, but it does not lie in $\operatorname{Code}(f)$. This contradicts preservation of the code below $V_\lambda$. Hence $j(f)(\kappa)=f(\kappa)$. Set $\theta:=j(f)(\kappa)$. Then $\theta=f(\kappa)<\lambda<\delta$, so $f(\kappa)\leq\theta$, $j(\kappa)>\lambda>\theta$, and $V_\theta\subset V_\lambda\subset M$. Restricting the equality $j(A)\cap V_\lambda=A\cap V_\lambda$ to $V_\theta$ gives $j(A)\cap V_\theta=A\cap V_\theta$. Together with $f``\kappa\subset\kappa$, this proves clause 2. [/step]

[step:Use a least-failure function to rule out failure of strongness]Assume clause 2. Fix $A\subset V_\delta$. Suppose, toward a contradiction, that clause 1 fails for this $A$. Thus, for every cardinal $\kappa<\delta$, there is an ordinal $\lambda$ with $\kappa<\lambda<\delta$ such that $\kappa$ is not $\lambda$-$A$-strong. Define a function $f_A:\delta\to\delta$ as follows. If $\xi<\delta$ is a cardinal and there is an ordinal $\lambda$ with $\xi<\lambda<\delta$ such that $\xi$ is not $\lambda$-$A$-strong, then $f_A(\xi)$ is the least such ordinal. If $\xi$ is not in that case, set $f_A(\xi):=\xi+1$. The values are below $\delta$: in the least-failure case this is part of the definition, and in the fallback case it follows from the fact that $\delta$ is a limit ordinal. By the class-embedding convention in the statement, $f_A$ is a set function on $\delta$. Apply clause 2 to the set $A$ and the function $f_A$. We obtain a cardinal $\kappa<\delta$, a transitive class $M$, and an elementary embedding $j:V\to M$ such that $\operatorname{crit}(j)=\kappa$, $f_A``\kappa\subset\kappa$, $\theta:=j(f_A)(\kappa)<\delta$, $f_A(\kappa)\leq\theta$, $j(\kappa)>\theta$, $V_\theta\subset M$, and $j(A)\cap V_\theta=A\cap V_\theta$. Since clause 1 was assumed to fail for $A$, the input $\kappa$ falls into the least-failure case. Hence $f_A(\kappa)$ is an ordinal satisfying $\kappa<f_A(\kappa)<\delta$, and $\kappa$ is not $f_A(\kappa)$-$A$-strong. On the other hand, the embedding $j$ witnesses that $\kappa$ is $\theta$-$A$-strong: its critical point is $\kappa$, it satisfies $j(\kappa)>\theta$, it has $V_\theta\subset M$, and it preserves $A$ through $V_\theta$. Because $f_A(\kappa)\leq\theta$, the same embedding also witnesses that $\kappa$ is $f_A(\kappa)$-$A$-strong. Indeed $j(\kappa)>\theta\geq f_A(\kappa)$, $V_{f_A(\kappa)}\subset V_\theta\subset M$, and restricting $j(A)\cap V_\theta=A\cap V_\theta$ to $V_{f_A(\kappa)}$ gives \begin{align*} j(A)\cap V_{f_A(\kappa)}=A\cap V_{f_A(\kappa)}. \end{align*} This contradicts the definition of $f_A(\kappa)$ as a failure level.[/step]

[guided]We assume clause 2 and fix a set $A\subset V_\delta$. To prove clause 1 for this $A$, suppose the opposite: every cardinal below $\delta$ fails to be $\lambda$-$A$-strong at some level below $\delta$. This lets us define a least-failure function $f_A:\delta\to\delta$. At a cardinal $\xi<\delta$, the value $f_A(\xi)$ is the least ordinal $\lambda$ with $\xi<\lambda<\delta$ such that $\xi$ is not $\lambda$-$A$-strong. At non-cardinal inputs, or at inputs where no such failure exists, we set $f_A(\xi)=\xi+1$. The class-embedding convention in the statement is exactly the hypothesis needed to ensure that this least-failure rule defines a legitimate set function on $\delta$. Now apply clause 2 to this particular function $f_A$ and to the same parameter $A$. Clause 2 gives a cardinal $\kappa<\delta$ and an elementary embedding $j:V\to M$ into a transitive class such that \begin{align*} \operatorname{crit}(j)=\kappa. \end{align*} It also gives, for $\theta:=j(f_A)(\kappa)$, \begin{align*} f_A(\kappa)\leq\theta<\delta. \end{align*} The embedding satisfies \begin{align*} j(\kappa)>\theta. \end{align*} It satisfies \begin{align*} V_\theta\subset M. \end{align*} Finally, it preserves the parameter $A$ through $V_\theta$: \begin{align*} j(A)\cap V_\theta=A\cap V_\theta. \end{align*} The point of adding the comparison $f_A(\kappa)\leq\theta$ to clause 2 is that it avoids any appeal to an invalid reflected parameter $j(A)$. We do not need to interpret $j(f_A)(\kappa)$ as a least failure for $j(A)$. We only use the ordinary value $f_A(\kappa)$ in the ground universe, where it was defined as the first level at which $\kappa$ fails to be $A$-strong. Since clause 1 was assumed to fail for $A$, the cardinal $\kappa$ is in the least-failure branch of the definition of $f_A$. Thus $f_A(\kappa)$ is a genuine failure level: $\kappa<f_A(\kappa)<\delta$, and $\kappa$ is not $f_A(\kappa)$-$A$-strong. But the embedding $j$ witnesses strongness at the larger level $\theta$. Because $f_A(\kappa)\leq\theta$, the witness restricts downward: $j(\kappa)>\theta\geq f_A(\kappa)$, $V_{f_A(\kappa)}\subset V_\theta\subset M$, and the equality $j(A)\cap V_\theta=A\cap V_\theta$ restricts to \begin{align*} j(A)\cap V_{f_A(\kappa)}=A\cap V_{f_A(\kappa)}. \end{align*} Therefore the same embedding $j$ witnesses that $\kappa$ is $f_A(\kappa)$-$A$-strong. This contradicts the definition of $f_A(\kappa)$ as a failure level. Hence the assumption that clause 1 fails for $A$ is impossible.[/guided]

[step:Combine the three implications] We have proved clause 1 implies clause 3, clause 3 implies clause 2, and clause 2 implies clause 1. Therefore the three displayed conditions are equivalent. [/step]