$\mathbb{P}(A|B)+\mathbb{P}(A^c|B^c)=1$ is false in general 
Let $A,B$ be some events in probability space $\Omega$. Show that
$$\mathbb{P}(A|B)+\mathbb{P}(A^c|B^c)=1$$ is false in general.

I thought of giving an example there events $A$ and $B$ are independent but then I get the correct equation because $$\mathbb{P}(A|B)+\mathbb{P}(A^c|B^c)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}+\frac{\mathbb{P}(A^c \cap B^c)}{\mathbb{P}(B^c)}=\mathbb{P}(A) + \mathbb{P}(A^c)=1.$$
My next idea was to check all other cases like when $A \subset B$ or when $A=B$ but everything seems to work out well. Can you help me of thinking of an example?


$$P(A|B)+P(A^c|B^c)=1\\\iff
\frac c{c+d}+\frac a{a+b}=1\\\iff
2ac+bc+ad=ac+ad++bc+bd\\\iff
ac=bd\\\iff
A \text{ and } B \text{ are }\textbf{independent events}.$$
The last equivalence is from my previous derivation here.
Therefore, the given statement is not in general true.
P.S. To say that the given statement is “false in general” may suggest that it cannot be true, which is not quite correct.
P.P.S. I wrote this related Answer, to another Question asked around the same time.
