How was this boolean expression further simplified?

Question

(ab+cd)(a'b'+c'd') = 1+ abc'd' + a'b'cd +1

so I'm stuck at

abc'd'+a'b'cd 

but the final answer is

(a+b)(c+d)+(a'+b')(c'+d')

What am I missing?

Answer 1

It seems to me that those two expressions are complementary, i.e. the only two cases where (a+b)(c+d)+(a'+b')(c'+d') are false are abc'd' and a'b'cd.

Edit: Somewhere along the line I think you've lost a ' and you're actually looking for one of these:

((ab+cd)(a'b'+c'd'))'
(ab+cd)'+(a'b'+c'd')'
((ab)'(cd)')+((a'b')'(c'd')')
(a'+b')(c'+d')+(a+b)(c+d)
(a+b)(c+d)+(a'+b')(c'+d')

(ab+cd)(a'b'+c'd')
(a'b'+c'd')(ab+cd)
((a+b)'+(c+d)')((a'+b')'+(c'+d')')
((a+b)(c+d))'((a'+b')(c'+d'))'
((a+b)(c+d)+(a'+b')(c'+d'))'

Answer 2

you cannot prove that (ab+cd)(a'b'+c'd') = (a+b)(c+d)+(a'+b')(c'+d') because it is not true.

take a=b=1, c=d=0:

(ab+cd)(a'b'+c'd') = (1+0)(0+1) = 1

but

(a+b)(c+d)+(a'+b')(c'+d') = (1*0)+(0*1) = 0

(assuming x' is "not")