I have to prove
ab + b'cd + acd = ab + b'cd
I figured it out using purely Boolean algebra but I also have to prove it by using Shannon's Expansion. But I'm not really sure how to use Shannons Expansion and then notes aren't very helpful.
I chose a, but I don't know how to choose the best term.

This is the best I could figure out:
f(a,b,c,d) = ab+ b'cd + acd = af(1,b,c,d) + a'f(0,b,c,d)
f(1,b,c,d) = 1b + b'cd +1cd
= b + b'cd + cd
= b + cd (b+1)
= b + cd

f(0,b,c,d) = 0b + b'cd + 0cd
= b'cd

f(a,b,c,d) = a(b + cd) + a'(b'cd)
= ab + acd +a'b'cd

Have I at least started this right? I don't know where to go from here.
Thanks

 

whoops I think I figured it out:
f(a,b,c,d) = a(b + cd) + a'(b'cd)
= ab + acd +a'b'cd
= ab+ abcd + ab'cd + a'b'cd (combining)
= ab * 1 + abcd + ab'cd + a'b'cd (identity)
= ab (1 + cd) + b'cd (a + a') (distributivity?) forgot which rule
= ab (1) + b'cd (1) (Null element and complement)
= ab + b'cd

Is that correct?

Also for the step
= ab+ abcd + ab'cd + a'b'cd (combining)
can i just go to
= ab + b'cd using the covering theorem?

Thanks for all the help