In what situations is this true: $P(A|B) = 1 - P(A|B^c)$?

Question

I am working on some software that uses Bayes' rule to find $P(B|A)$

It asks the user to define $P(A|B)$ but it says that $P(A|B^c)$ is optional, if the user chooses not to define it then it sets $P(A|B^c) = 1 - P(A|B)$.

This seems slightly odd to me. Is there a group of cases where $P(A|B^c) = 1 - P(A|B)$ holds true.

Related but does not answer: Does $P(A\mid B)$ determine $P(A\mid \bar{B})$?

Related: Does $P(A|B)+P(A|\bar{B}) \neq 1?$

Answer 1

See Does $P(A|B)+P(A|\bar{B}) \neq 1?$

as said in this answer by @drhab

$P(A|A)+P(A|A¯)=1+0=1$

So if $P(A)=P(B)$