by [@DG 5.25] is 16, so there must be at least one more linearly independent relation. Where else can we find it? It is easy to verify that all the seven RDT (including four FDT) can be derived from RDS. However, we know that a product of two weight two MPVs is of weight four. So on each of the five RDS (including two FDS) in $\MPV(2,3)$ we can multiply any one of the nine MPVs of $(w,N)=(2,3)$ to get a relation in $\MPV(4,3)$. For instance, we have a FDS $$Z(y_{1,1}\ast y_{1,1}-y_{1,1}\sha y_{1,1})
    =L_3(2|2)+2L_3(1,1|1,1)-L_3(1,1|1,0)=0.$$ Multiplying by $L_3(1,1|1,1)=Z(y_{1,1}y_{1,2})$ we have a new relation not derivable from RDS in $\MPV(4,3)$: $$\begin{aligned}
 & Z\big(y_{1,1}y_{1,2} \sha[y_{2,0}+2y_{1,1}y_{1,2}-2y_{1,1}y_{1,0}] \big)\\
=&L_3(1,1,2|1,1,0)+2L_3(1,2,1|1,1,0)+2L_3(2,1,1|1,1,0)+L_3(2,1,1|2,2,1)
+4L_3(\{1\}^4|1,1,2,1)\\
+&8L_3(\{1\}^4|1,0,1,0)-6L_3(\{1\}^4|1,0,0,1)
-4L_3(\{1\}^4|1,0,1,2)-2L_3(\{1\}^4|1,1,2,0)=0.\end{aligned}$$ Such relations coming from the lower weights are called *lifted relations (from lower weights).* In this way, when $(w,N)=(4,3)$ we can produce 45 lifted RDS relations from weight two, 58 from weight three. We may also lift RDT and obtain nine and six relations from weight two and three, respectively. However, all the lifted relations together only produce one new linearly independent relation, as expected. Hence we find totally 128 linearly independent relations among the 144 MPVs of $(w,N)=(4,3)$. This implies that $d(4,3)\le 