how to prove bc + a'cd + ab'cd + bd' + bc'd = cd + b with Boolean algebra? I'm trying to prove this equation using Boolean algebra : (it is a digital design problem)
$$bc + \bar acd + a \bar bcd + b \bar d + b \bar cd = cd + b$$
What I'm done:
$$bc + \bar acd + a \bar bcd + b \bar d + b \bar cd$$
$$=cd(\bar a + a \bar b) + b(c + \bar cd + \bar d)$$
from this I think both $(\bar a + a \bar b)$  and $(c + \bar cd + \bar d)$ must be $1$.
so I tried to multiply this to something equal to $1$ to simplify it:
$$(c + \bar cd + \bar d)$$
$$=(c + \bar cd + \bar d)(c + \bar c)$$
$$=(c + \bar cd + c \bar d + \bar c \bar d)$$
$$=c(1+d) + \bar c(d+ \bar d)$$
$$=(c+ \bar c)$$
$$=1$$
but for the second term I tried the same above with $(a + \bar a)$ , $(b + \bar b)$ but none of them worked. so what is the trick here?
 A: 
$$=cd(\bar a + a \bar b) + b(c + \bar cd + \bar d)$$
from this I think both $(\bar a + a \bar b)$  and $(c + \bar cd + \bar d)$ must be $1$.

Given that your goal is $cd + b$ that makes a lot of sense, but it is not true. Indeed, $\bar a + a \bar b \not = 1$
How can that be?  Well, all you really have shown is that your original expression simplifies to
$$cd(\bar a + a \bar b) + b(c + \bar cd + \bar d)$$
and that would equal $cd + b$ if it were true that $(\bar a + a \bar b)$  and $(c + \bar cd + \bar d)$ are both $1$
which is not the same as saying that must be true that $(\bar a + a \bar b)$  and $(c + \bar cd + \bar d)$ are both $1$
Indeed, the fact that it is not true that $(\bar a + a \bar b)$  and $(c + \bar cd + \bar d)$ are both $1$ does not mean that $cd(\bar a + a \bar b) + b(c + \bar cd + \bar d)$ does not simplify to $cd + b$.  It does simplify to that ... you just need to try a different route.
In sum: Some proof strategies work, others don't. Your particular strategy didn't. So try a different one.
I suggest you use:
Reduction
$p + \bar p q = p + q$
(Proof: $p + \bar p q = (p + \bar p)(p +q) = 1(p+q)=p + q$)
Or even:
Generalized Reduction
$pr + \bar p q r = pr + qr$
(Proof: $pr + \bar p q r = (p+ \bar p q)r = (p + q)r = pr + qr$
And of course:
Absorption
$p + pq = p$
(Proof: $p + pq = p(1+q)=p1=p$)
and:
Adjacency
$pq+p\bar q = p$
(Proof: $pq+p\bar q = p(q + \bar q) = p1 = p$)
With that:
$$bc + \bar acd + a \bar bcd + b \bar d + b \bar cd = \text{ (Gen'd Reduction x 2)}$$
$$bc+ \bar acd + a \bar bcd + b \bar d + b= \text{(Absorption x 2)}$$
$$\bar acd + a \bar bcd + b= \text{(Reduction)}$$
$$\bar acd + a cd + b= \text{(Adjacency)}$$
$$cd + b$$
