I was reading the following from a book of probability theory:

> 
> A contradiction $\bar{A}A$ implies all propositions, true and false. (Given any two propostions $A$ and $B$, we have $A\Rightarrow(A+B)$, therefore $\bar{A}A\Rightarrow\bar{A}(A+B)=\bar{A}A+\bar{A}B\Rightarrow B$.)
> 
> 
> 

I understand these deductions except the last one. If $\bar{A}A$ is assumed to be true, then the validity of $\bar{A}B$ should be undetermined merely from $\bar{A}A+\bar{A}B$. Because whether $\bar{A}B$ is true or not, $\bar{A}A+\bar{A}B$ is true anyways. Am I right?