Don't Booley me

Let's convert this statement into Boolean algbera:

$(b+\color{#D61F06}{\overline{c}})(\overline{a}b+\color{#D61F06}{a}+\overline{b}c)+\overline{a}+ac$

Using the distributive law we see that there will be a $\color{#D61F06}{a\overline{c}}$ term in the expansion. Since

$a\overline{c}+\overline{a}+ac = a(\overline{c}+c)+\overline{a} = a+\overline{a} = 1$

and $p+1 = 1$ for any proposition $p$ , we get that the statement given is a tautology!

Let there be a falsifying assignment. Then a must be assigned as 1 otherwise

                    ... +~a+ ac

will be true

Now for a.c to be false, c must be 0.

So, in (b+~c), ~c is true. In (~a.b + a + ~b.c), a is true. Hence (b+~c).(~a.b + a + ~b.c) is true.

So there can be no falsifying assignment. Hence the proposition is a tautology.

There is no need to solve the entire expression.Here are few steps for easy evaluation

1. Consider,

(b V ~c) ^(and) with remaining ((~a^b)VaV(~b^c))

whatever may be the RIGHT PART of '^' just neglect and MAKE (b V~c) as ZERO(FALSE) i.e b=0,c=1.

    Thus   when (and)ed with right part gives zero.

1. Consider,

   (derived from previous step):- (FALSE)0 V ~a V(a^c)

already c=1, choose whatever values for 'a' it will yield TRUE always .