How to show these two expressions are equivalent using Boolean Algebra?

Question

I have two boolean functions:

$$f(a, b, c, d) = \bar{a}cd + a\bar{c}d + ab\bar{d} + abc$$ and $$g(a, b, c, d) = d(a \oplus c) + ab$$

I know these two functions are equivalent because I tried every combination by using a truth table.

Now, can somebody help me show the equivalence of these two expressions by using basic boolean algebra laws?

How are these equivalent, precisely? Pick $d=c=0,$ then $f\ne g,$ since $0 \ne ab$ in general. — Chickenmancer, Jun 25 '18 at 01:15
There is a term $ab\bar{d}=ab$ (for $\bar d=1$) in the first expression — Andrei, Jun 25 '18 at 01:17
@Chickenmancer When $a=b=1$ and $d=c=0$, $ab=ab\bar{d}=1$; so both $f=g=1$ — Truth-seek, Jun 25 '18 at 01:19

score 0 · Accepted Answer · answered Jun 25 '18 at 02:00

$\def\onot#1{\overline{#1}}$Start with $g$, use the definition for $\oplus$ (exclusive or), and tautologies $a+\onot a$. Then use distribution, commutation, association, and idemptence.

$$\begin{split} g(a,b,c,d)&= d(a\oplus c)+ab \\ &=(\onot ac+a\onot c)(b+\onot b)d+ ab(c+\onot c)(d+\onot d)\\&~~\vdots\\ & =\onot acd+a\onot cd+ab\onot d+abc\\ & = f(a,b,c,d)\end{split}$$

How to show these two expressions are equivalent using Boolean Algebra?

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