The 2-primary Part of Ш

I was reading Silverman's Arithmetic of Elliptic Curves I have a question on computing the Mordell-Weil group of an elliptic curve over $E(\mathbb{Q})$.

Adapting the argument given in Silverman we have that we successive compute $$S^{(2)}(E/\mathbb{Q}) = S^{(2, 1)}(E/\mathbb{Q}) \supset S^{(2, 2)}(E/\mathbb{Q}) \supset S^{(2, 3)}(E/\mathbb{Q}) \supset \ldots$$ and $$T_{(2,1)}(E/\mathbb{Q}) \subset T_{(2,2)}(E/\mathbb{Q}) \subset T_{(2,3)}(E/\mathbb{Q}) \subset \ldots$$ where $T_{(2,r)}(E/\mathbb{Q})$ generated by the set $$W = \{ P \in E(\mathbb{Q}) \; | \; h(P) \leq r \}.$$

Why is it that once we have that $$S^{(2, m)} (E/\mathbb{Q}) = T_{(2, r)} (E/\mathbb{Q}),$$ then $2^{m-1}Ш(E/\mathbb{Q})[2^m] = 0$?

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The Selmer groups $S^{(2,m)}$ get smaller as $m$ increases, but they all contain the image of $E(K)/2E(K)$ in the Selmer group $S^{(m)}(E/K)$ (see Proposition 4.12 of Chapter X in Silverman, and the diagram immediately above it). In other words, $S^{(2,m)}$ approaches $E(K)/2E(K)$ "from above" as $m$ increases.
On the other hand, the $T_{(2,r)}$ groups get larger as $r$ increases, and they consist of actual points of $E(K)$, hence their images in $E(K)/2E(K)$ are all contained in the Selmer group $S^{(2)}(E/K)$. So $T_{(2,r)}$ approaches $E(K)/2E(K)$ "from below" as $r$ increases.
So if ever you can show that $S^{(2,m)} = T_{(2,r)}$ for some $m$ and $r$, then you have squashed the upper and lower bounds for $E(K)/2E(K)$ together, hence identified the group $E(K)/2E(K)$, which is the goal of this entire chapter of Silverman. For such a value of $m$, we have $E(K)/2E(K) = S^{(2,m)}(E/K)$, so the quotient $2^{m-1}Ш(E/\mathbb{K})[2^m]=0$ (again, see the exact sequence in Proposition 4.12).