If I have the boolean equation:
H = M'CD' + MC + MC' + CRD
I think I can combine so that it's
H = M'CD' + M(C + C') + CRD
Does C + C' go to simplify to zero? So, I'm left with
H = M'CD' + CRD
If I have the boolean equation:
H = M'CD' + MC + MC' + CRD
I think I can combine so that it's
H = M'CD' + M(C + C') + CRD
Does C + C' go to simplify to zero? So, I'm left with
H = M'CD' + CRD
Karnaugh maps are your best friend. Otherwise, observe that:
$$ \begin{align*} H &= M'CD' + M + CRD \\ &= M'CD' + M(1) + CRD \\ &= M'CD' + M(CD'+1) + CRD \\ &= M'CD' + MCD' + M + CRD \\ &= (M' + M)CD' + M + CRD \\ &= (1)CD' + M + CRD \\ &= CD' + M + CRD \\ &= CD' + CRD + M \\ &= C(1)D' + CRD + M\\ &= C(1 + R)D' + CRD + M\\ &= CD' + CRD' + CRD + M\\ &= CD' + CR(D' + D) + M\\ &= CD' + CR(1) + M\\ &= CD' + CR + M\\ \end{align*} $$