$\mathbb{P}(A|B)+\mathbb{P}(A^c|B^c)=1$ 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?

What is true is $$P(A\mid B)+P(A^c\mid B)=1.$$
If your equality was true, it would mean $$P(A^c\mid B)=P(A^c\mid B^c),$$ which is the definition of

$A^c$ and $B$ are independent.

That can be shown to be equivalent to

$A$ and $B$ are independent.

So your equality is true if and only if  $A$ and $B$ are independent.
But for example, if $A$ and $B$ are the same event, then $$P(A\mid A)+P(A^c\mid A^c)=2.$$
Then your equality is not true.
If $B=A^c$ then you get $$P(A\mid A^c)+P(A^c\mid A)=0.$$
So your sum can really be anywhere between $0$ and $2.2
