Decryption of message in IBE without random oracle using bilinear pairing registration? I find the following IBE scheme from the videos posted and i don't understand the decryption algorithm, will any one please elaborate  the  6th step

scheme


Setup($\lambda$)  :     $(\mathbb{G}, \mathbb{G}_T, g, P)$   &lt;-- GenBilgroup($\lambda$)

$\alpha\leftarrow_R\mathbb{F}_p$
Public Parameters :   $pp = [ g , y = g^\alpha  , g_1, h ] \in \mathbb{G}$
Master secret : $mk = g_1^\alpha $
Keygeneration:  $k(mk,id)$: $d_1= mk\cdot(Y^{id\cdot h})^r, d_2 = g^r $
where   $r\leftarrow_R\mathbb{F}_p $
Encryption: $E(pp,id,m)$ :  $c_1 = g^s, c_2=(Y^{id\cdot h})^s, c_3=m\cdot e(y, g_1)^s$
Decryption: $e(c_1,d_1)/e(c_2,d_2)$ will give $e(y,g_1)^s$  How?
             and $m$ is obtained by dividing $c_3$ by that value.
I don't know whether it is correct

$$e(c_1,d_1) = e(g^s, mk\cdot(y^{id\cdot h})^r)$$

$$=e(g^s,mk)\cdot e(g^s, (y^{id\cdot h})^r)$$

$$= e(g^s , g_1^\alpha ) \cdot e(g^r , (y^{id\cdot h})^s)$$

$$= e(y , g_1^\alpha ) \cdot e(g^r , (y^{id\cdot h})^s)$$

$$e(c_2,d_2)  = e((Y^{id\cdot h})^s , g^r)$$

$$= e((Y^{id\cdot h})^s , g^r)$$

therefore $e(c_1,d_1)/e(c_2,d_2) =  e(y , g_1^\alpha )$