yes!
COMPLEMENT OF DUAL=DUAL OF COMPLEMENT
Complement of dual
we have function,$F(w,x,y,z,+,.,0,1)$
dual of it, $F_D(w,x,y,z,.,+,0,1)$
complement of it,$[F_D(w,x,y,z,.,+,0,1)]' =F_N(w',x',y',z',+,.,1,0)$
Dual of complement
we have function,$F(w,x,y,z,+,.,0,1)$
complement of it, $[F(w,x,y,z,+,.,0,1)]' =F_C(w',x',y',z',.,+,1,0)$
dual of it,$F_N(w',x',y',z',+,.,1,0)$
Example:
consider a + b' .
i)( COMPLEMENT OF DUAL)its dual is ab'. now the complement is (ab')'=a'+b
ii)(DUAL OF COMPLEMENT)its complement is a'b. dual would be a'+b.