Suppose we have a type-3 pairing $(q,\mathbb{G_1},\mathbb{G_2},\mathbb{G_T},g_1,g_2,e)$, that holds $SXDH$ assumption. Given $(g_1,g_1^a,g_1^b,g_1^c,g_1^m,g_1^{m'},g_1^{a\cdot m}, g_1^{b\cdot m'},g_1^{s}, g_2,(g_2^{a\cdot r},g_2^{b\cdot r}),(g_2^{b\cdot r'},g_2^{c\cdot r'}))$ to determine whether or not $s=m\cdot c $, where $a,b,c,m,m',r,r' \leftarrow \mathbb{Z}^*_q$.

UPDATA-1: I find a assumption that shows above problem is hard (this assumption called SAXDH): OXT-Article

How does it show that SAXDH assumption implies SXDH assumption?

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