Need to be sure of this boolean algebra problem's solution

[Amr719]

Homework Statement

Express the function Y= (abd + c)' + ((acd)'+(b)')' as the complete disjunctive normal form:
2.1 by applying Boole's theorerm,

Homework Equations

The Attempt at a Solution

I separated the equations to two terms (T1,T2)

T1= (abd + c)' T2=((acd)'+(b)')'

T1= (abd+c)' T2=((acd)'+(b)')'
=(abd)'.(c)' =(acd)".(b)''
=((a)'+(b)'+(c)'+(d)'). (c)' = abcd
= a'c'.(b+b') + b'c'(a+a') + c'd'(a+a')
=a'c'(bd+(bd)')+b'c'(ad+a'd')+c'd'(ab+a'b')
=a'c'bd+a'c'b'd'+b'c'ad+a'd'b'c'+abc'd'+a'b'c'd'T1+T2= a'bc'd+ab'c'd+abc'd'+abcd

 

[haruspex]

1= (abd+c)' T2=((acd)'+(b)')'
=(abd)'.(c)' =(acd)".(b)''
=((a)'+(b)'+(c)'+(d)'). (c)' = abcd

Not sure how to interpret your working. I think you intended this as two columns of working, the left hand for T1 and the right hand for T2. In which case, your last steps in what I quoted above are
=(abd)'.(c)' =((a)'+(b)'+(c)'+(d)')
And
=(acd)".(b)'' = abcd
Neither of those are correct. Take them in smaller steps.