We have learned that the logical implication $(C=>D)$ is equivalent to the following expressions,

$\frac{C}{b}ar(D)$ is false $,$

$\frac{?}{\text{b}\text{a}\text{r}}(C)+D$ is true $,$

$C=CD,$

Now, using the above expressions, show that if$,$

$(B=\frac{?}{b}>ar(A)),$

is True then,

$(A\frac{+}{b}ar(B))(\frac{?}{\text{b}\text{a}\text{r}}(A)+\frac{A}{b}ar(B))\frac{+}{b}ar(A)B(A+B),$

is also True. In other words, the above two expressions are equivalent to each other.