= Solution
The <Clebsch-Gordan decomposition for SU2> states
$$
\boxed{V_m\otimes V_n\cong\bigoplus_{r=0}^{\min(m,n)}V_{m+n-2r}.}
$$
Realize the <tensor product> as <polynomials> of bidegree $(m,n)$ in $(x_1,y_1)$ and $(x_2,y_2)$. The <determinant> $D=x_1y_2-y_1x_2$ is invariant under the simultaneous SU2 action. For each $r$, the vector
$$
v_r=D^r x_1^{m-r}x_2^{n-r}
$$
has <torus> weight $q=m+n-2r$ and is killed by $E=x_1\partial_{y_1}+x_2\partial_{y_2}$. Applying $F=y_1\partial_{x_1}+y_2\partial_{x_2}$ produces $q+1$ nonzero vectors of distinct weights, with $F^{q+1}v_r=0$. The relations $[E,F]=H$ and $H v_r=qv_r$ give $E F^jv_r=j(q-j+1)F^{j-1}v_r$, so their span is an irreducible copy of $V_q$.
These irreducible submodules have distinct <highest weights> and form a <direct sum>: a sum of earlier ones cannot contain an <irreducible module> inequivalent to all its summands. Equivalently use complete reducibility, obtained by averaging an <inner product> over the compact group. Assuming $m\leq n$, their total dimension is
$$
\sum_{r=0}^m(m+n-2r+1)=(m+1)(n+1),
$$
which equals the dimension of the <tensor product>. Hence they exhaust it, proving the theorem.
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