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Write $A = [A_1 ~~ A_2]$ with $A_1 \in \mathbb{Z}_q^{n\times m'}$ and $A_2 \in \mathbb{Z}_q^{n\times (m-m')}$. Likewise, $e = (e_1 ~~ e_2)$ with $e_1 \in \mathbb{Z}_q^{m'}$ and $e_2 \in \mathbb{Z}_q^{m-m'}$. Then, $$Ae = 0 \bmod q \iff A_2e_2 = -A_1e_1 \bmod q.$$ So, given an instance of SIS, that is, an $n\times m$ matrix $A$, if you have an oracle to solve ...