Boolean Algebra logic simplification

Question

By applying the conventional 12 rules of simplification i am unable to simplify this expression! objective is to simplify the expression to make it as contract as possible and can easily be implemented.

A.B'.C + B.C + A.C'

Answer 1

First method to solve it using Boolean algebra:

A.B'.C + B.C + A.C'

C.(B + A.B') + A.C' (Take C as common factor)

C.(B + A)(B+B') + A.C' (Using Distributive Law)

C.(B + A).1 + A.C'

B.C + A.C + A.C'

B.C + A.(C + C') (Take A as common factor )

B.C + A.1

A + B.C

second method is using k-map

Answer 2

Similarly to Abdul's proof:

A.B'.C + B.C + A.C' = (A.B'+ B).C + A.C'   (common factor C)
                    = (A + B).C + A.C'     (see below)
                    = A.C + B.C + A.C'     (distribute C)
                    = A + B.C              (A.C + A.C' = C, right?)

Why A.B' + B = A + B?

A.B' + B = A.B' + A.B + B (because B = A.B + B since B includes A.B)
         = A.(B'+ B) + B  (common factor A)
         = A + B