[proofplan] We prove the two implications separately. The supercompactness implication follows directly from the formalized definition: a supercompactness embedding for a target ordinal $\lambda$ already has the properties required of a strength embedding for that same $\lambda$. For the implication from strength to measurability, we take an elementary embedding witnessing enough strength to have $j(\kappa) > \kappa$, derive from it the ultrafilter $U = \{A \subset \kappa : \kappa \in j(A)\}$, and verify directly that $U$ is a nonprincipal $\kappa$-complete normal ultrafilter on $\kappa$. [/proofplan] [step:Extract the strength witnesses from supercompactness] Assume that $\kappa$ is supercompact. Let $\lambda$ be an ordinal. By the definition of supercompactness in the theorem statement, there exist a transitive class $M$, closed under $\lambda$-indexed sequences from $V$, and a nontrivial elementary embedding \begin{align*} j: V \to M \end{align*} whose critical point is $\operatorname{crit}(j) = \kappa$, such that $j(\kappa) > \lambda$ and $V_\lambda \subset M$. The definition of strength asks only for a transitive class $M$ and such an elementary embedding with $\operatorname{crit}(j) = \kappa$, $j(\kappa) > \lambda$, and $V_\lambda \subset M$; the sequence-closure clause is extra structure and is not needed for strength. Since this holds for every ordinal $\lambda$, $\kappa$ is strong. This proves the first implication. [/step] [step:Derive an ultrafilter from an elementary embedding witnessing strength] Assume that $\kappa$ is strong. Here $V$ denotes the universe of sets, and for an elementary embedding $j$ the symbol $\operatorname{crit}(j)$ denotes the least ordinal moved by $j$. Applying the definition of strength with target ordinal $\kappa$, there is a transitive class $M$ and an elementary embedding \begin{align*} j: V \to M \end{align*} such that $\operatorname{crit}(j) = \kappa$, $j(\kappa) > \kappa$, and $V_\kappa \subset M$. The inclusion $V_\kappa \subset M$ is part of the strength witness, but the construction below uses only the embedding, its critical point, and the inequality $j(\kappa) > \kappa$. Define \begin{align*} U := \{A \subset \kappa : \kappa \in j(A)\}. \end{align*} Thus $U$ is a collection of subsets of $\kappa$. We first show that $U$ is an ultrafilter on $\kappa$. Let $A \subset \kappa$. Since $j$ is elementary and fixes all ordinals below $\kappa$, $j(\kappa)$ is an ordinal strictly above $\kappa$, and $j(A) \subset j(\kappa)$. Also \begin{align*} j(\kappa \setminus A) = j(\kappa) \setminus j(A). \end{align*} Therefore exactly one of $\kappa \in j(A)$ and $\kappa \in j(\kappa \setminus A)$ holds. Hence exactly one of $A \in U$ and $\kappa \setminus A \in U$ holds. We also verify the filter axioms. Since $\kappa \in j(\kappa)$, we have $\kappa \in U$, and since $j(\varnothing) = \varnothing$, we have $\varnothing \notin U$. If $A, B \in U$, then $\kappa \in j(A)$ and $\kappa \in j(B)$, so \begin{align*} \kappa \in j(A) \cap j(B) = j(A \cap B). \end{align*} Thus $A \cap B \in U$. If $A \in U$ and $A \subset B \subset \kappa$, then elementarity gives $j(A) \subset j(B)$, and $\kappa \in j(A)$ implies $\kappa \in j(B)$. Hence $B \in U$. Together with the complement decision property, these filter axioms show that $U$ is an ultrafilter. For each $\alpha < \kappa$, elementarity and $\operatorname{crit}(j) = \kappa$ give \begin{align*} j(\{\alpha\}) = \{\alpha\}. \end{align*} Since $\kappa \ne \alpha$, we have $\kappa \notin j(\{\alpha\})$, so $\{\alpha\} \notin U$. Thus $U$ is nonprincipal. [guided] The purpose of the embedding is to turn the point $\kappa$ inside the target model into a test for largeness of subsets of $\kappa$. We define \begin{align*} U := \{A \subset \kappa : \kappa \in j(A)\}. \end{align*} This definition makes sense because $j(A)$ is a subset of $j(\kappa)$ whenever $A \subset \kappa$, and $j(\kappa) > \kappa$, so $\kappa$ is an element of the ambient ordinal $j(\kappa)$. To prove that $U$ is an ultrafilter, we first verify the ordinary filter axioms. The whole set belongs to $U$ because $\kappa \in j(\kappa)$, while the empty set does not belong to $U$ because $j(\varnothing) = \varnothing$. If $A, B \in U$, then $\kappa \in j(A)$ and $\kappa \in j(B)$, so \begin{align*} \kappa \in j(A) \cap j(B) = j(A \cap B). \end{align*} Therefore $A \cap B \in U$. If $A \in U$ and $A \subset B \subset \kappa$, then elementarity gives $j(A) \subset j(B)$. Since $\kappa \in j(A)$, it follows that $\kappa \in j(B)$, and hence $B \in U$. Thus $U$ is a proper filter on $\kappa$. It remains to prove maximality of this filter. Take any subset $A \subset \kappa$. The complement of $A$ in $\kappa$ is $\kappa \setminus A$. Since $j$ is elementary, it preserves the operation of taking complements inside the relevant ordinal: \begin{align*} j(\kappa \setminus A) = j(\kappa) \setminus j(A). \end{align*} Because $j(A) \subset j(\kappa)$, the ordinal $\kappa$ lies in exactly one of the two complementary subsets $j(A)$ and $j(\kappa) \setminus j(A)$. Therefore exactly one of $A$ and $\kappa \setminus A$ belongs to $U$. This is the ultrafilter decision property, so the proper filter $U$ is an ultrafilter. Now we check nonprincipality. Let $\alpha < \kappa$. Since $\operatorname{crit}(j) = \kappa$, the embedding fixes every ordinal below $\kappa$, so $j(\alpha) = \alpha$. Hence \begin{align*} j(\{\alpha\}) = \{\alpha\}. \end{align*} But $\kappa \ne \alpha$, so $\kappa \notin \{\alpha\} = j(\{\alpha\})$. Thus $\{\alpha\} \notin U$ for every $\alpha < \kappa$. No singleton has measure one, so the ultrafilter is nonprincipal. [/guided] [/step] [step:Verify $\kappa$-completeness of the derived ultrafilter] Let $\delta < \kappa$, and let \begin{align*} \langle A_i : i < \delta \rangle \end{align*} be a sequence of elements of $U$. Since $\delta < \kappa = \operatorname{crit}(j)$, the embedding fixes $\delta$ and each index $i < \delta$. Applying $j$ to the sequence gives \begin{align*} j(\langle A_i : i < \delta \rangle) = \langle j(A_i) : i < \delta \rangle. \end{align*} Because $A_i \in U$ for every $i < \delta$, we have $\kappa \in j(A_i)$ for every $i < \delta$. Therefore \begin{align*} \kappa \in \bigcap_{i < \delta} j(A_i). \end{align*} Elementarity gives \begin{align*} j\left(\bigcap_{i < \delta} A_i\right) = \bigcap_{i < \delta} j(A_i). \end{align*} Hence $\kappa \in j(\bigcap_{i < \delta} A_i)$, so \begin{align*} \bigcap_{i < \delta} A_i \in U. \end{align*} Thus $U$ is $\kappa$-complete. [/step] [step:Verify normality of the derived ultrafilter] Let $S \in U$, and let \begin{align*} f: S \to \kappa \end{align*} be a regressive map, meaning that $f(\alpha) < \alpha$ for every nonzero $\alpha \in S$. Since $S \in U$, we have $\kappa \in j(S)$. By elementarity, $j(f): j(S) \to j(\kappa)$ is regressive on the nonzero elements of $j(S)$. Applying this at $\kappa \in j(S)$ gives \begin{align*} j(f)(\kappa) < \kappa. \end{align*} Define the ordinal \begin{align*} \beta := j(f)(\kappa). \end{align*} Then $\beta < \kappa$, so $j(\beta) = \beta$. Define the fiber \begin{align*} S_\beta := \{\alpha \in S : f(\alpha) = \beta\}. \end{align*} Since $\kappa \in j(S)$ and $j(f)(\kappa) = \beta = j(\beta)$, we have \begin{align*} \kappa \in j(S_\beta). \end{align*} Therefore $S_\beta \in U$. This proves that every regressive map on a $U$-large set is constant on a $U$-large subset, so $U$ is normal. [/step] [step:Conclude measurability and chain the implications] We have constructed a nonprincipal $\kappa$-complete normal ultrafilter $U$ on $\kappa$. By the definition of a measurable cardinal, $\kappa$ is measurable. Therefore every strong cardinal is measurable. Combining this with the first implication gives \begin{align*} \text{supercompact} \implies \text{strong} \implies \text{measurable}. \end{align*} This proves the theorem. [/step]