What is the difference between $P(A|B)$ and $P(\neg B | \neg A)$? If this is trivial or duplicate, I am very sorry.
What is the difference between $P(A|B)$ and $P(\neg B | \neg A)$ for events $A$ and $B$ with a probabiilty measrue $P$ on the events?
Are they equal? I cannot derive the above by algebra on probability
 A: Consider your favorite example of independent events $A$ and $B$.  With $A$ and $B$ independent, that implies that $\neg A$ and $\neg B$ are independent of one another as well.
As they are independent, we have $\Pr(A\mid B) = \Pr(A)$ and that $\Pr(\neg B\mid \neg A) = \Pr(\neg B)$
So, you are asking if given two events $A$ and $B$ whether or not $\Pr(A)$ is necessarily equal to $\Pr(\neg B)$.  It is clear that this does not need to be the case.
A: No, they are not equivalent.
Take, for example, a set of shapes containing one red square, 49 red circles, and 99 blue squares.
The probability that a shape is blue given it is a square is $99\%$.  The probability that a shape is a circle (not a square) given it is red (not blue) is $98\%$.
A: 
If $a,b,c, d$ are probabilities, $$P(X|Y)=\frac{c}{c+d}\not\equiv \frac{a}{a+d}=P(Y^\complement|X^\complement).$$

$$ \begin{array}{r}  \begin{array}{c|c|c}
  \style{font-family:inherit}{\text{time of complaint}\bigg\\ \text{reason for complaint}}  & \style{font-family:inherit}{\textbf E\text{lectrical}}  & \style{font-family:inherit}{\textbf M\text{echanical}}  & \style{font-family:inherit}{\textbf L\text{ooks}}
\\\hline
  \style{font-family:inherit}{\textbf D\text{uring guarantee period}}  & 18\%  & 13\%  & 32\%
\\[0pt]\hline
  \style{font-family:inherit}{\textbf A\text{fter guarantee period}}  & 12\%  & 22\%  & 3\%
 \end{array}\hskip-5.5pt  \end{array} $$
$$P(L|D)=\frac{32}{18+13+32}=50.8\%\neq52.3\%=\frac{12+22}{18+13+12+22}=P(D^\complement|L^\complement).$$
