0 votes
1
Following is the way of checking the one to one functionorCan i use bi implication in between these?If not,then why?
0 votes
2
L = {a^nb^m : n >= m+3}below context grammer is correct??S == aA | AaA == aAb | bAa | abA | baA | aa
1 votes
4
Prove or disprove: $\begin{align*} \log_8x = \frac{1}{2}.\log_{2}x \end{align*}$.