Probability of A given B complement

Question

In this other question it is laid out the following identity. $$ P(A|B^c) = 1 - P(A^c|B^c) $$ Been trying to prove it without success. I can only prove that $$ 1-P(A^c|B^c) = \frac{P(A)}{P(B^c)} $$ so I'm starting to think that identity on the other question is wrong. Can anyone help me prove if the first identity is true?

Edit: my result explanation

$$ 1 - P(A^c|B^c) = 1 - \frac{P(A^c \cap B^c)}{P(B^c)} $$ $$ 1 - \frac{P(A^c \cap B^c)}{P(B^c)}=1- \frac{1-P(A)-P(B)}{P(B^c)}=1-(\frac{1-P(B)}{P(B^c)} - \frac{P(A)}{P(B^C)}) = 1-(1-\frac{P(A)}{P(B^c)})= \frac{P(A)}{P(B^c)} $$

Answer 1

$P(A\mid B^c)$

$=\dfrac{P(A, B^c)}{P(B^c)}$

$=\dfrac{P(B^c) - P(A^c, B^c)}{P(B^c)}$

$=\dfrac{P(B^c)}{P(B^c)}-\dfrac{P(A^c, B^c)}{P(B^c)}$

$=1- P(A^c\mid B^c)$

Answer 2

$A \cup B=(A\setminus B) \cup B$ and $P(A \cup B)=P(A\setminus B)+P(B)$. This gives $1-P(A^{c}\cap B^{c})=P(A\setminus B)+P(B)$ or $1-P(B)+P(A^{c}\cap B^{c})=P(A\setminus B)$. Divide throughout by $1-P(B)$. Can you finish?

Answer 3

$P(A|B^C)+P(A^C|B^C)=1$ because both cases are incompatible, and if you know $B^C$ to be true, either $A$ happens or $A^C$ happens, which gives it a probability of 1.

Answer 4

Rhizome's answer is clear enough, but here is a diagram to show it:

Since $A$ and $A'$ are the only two possibilities for event $A$, $P(A | B') + P(A' | B') = P(B' | B') = 1$ by the law of total probability.