Show that if $P(B|A) =1$ then $P(A^C|B^C)=1$

Question

I want to show that if $P(B|A) =1$ then $P(A^C|B^C)=1$ where A and B are events with probabilities not equal to 0 or 1. The question comes from Blitzstein's Introduction to Probability (Chapter 2, question 17(a)) and, as a hint, notes that Bayes' rule and the Law of Total Probability should be applied. I thought the most direct way would be as follows:

\begin{align*} P(A^C|B^C) = 1 -P(A|B^C) && \text{Definition of complement}\\ P(A^C|B^C) = 1 - \bigg(\dfrac{P(B^C|A)P(A)}{P(B^C)}\bigg) && \text{Bayes' rule}\\ P(A^C|B^C) = 1 - \bigg(\dfrac{(1-P(B|A))P(A)}{P(B^C)}\bigg) && \text{Definition of complement}\\ P(A^C|B^C) = 1 - \bigg(\dfrac{(0)P(A)}{P(B^C)}\bigg) && \text{Using P(B|A) = 1}\\ \end{align*}

And thus, $P(A^C|B^C) = 1$. However, since I didn't at all use the Law of Total Probability as the question hinted at, I'm not confident whether this is another way of performing this proof or if I've done something wrong here. I would really appreciate any assistance in pointing out if I did something wrong.

Answer 1

The following answer is wrong, but is being left in (i.e. not deleted) as an interesting example of me going off the rails.

Alternative:

In general, when $R,S$ are both events with probability strictly between $0$ and $1$, exclusive, then $p(R|S) = 1$ is logically equivalent to $(S \implies R).$

Edit
Thanks to TonyK for indicating that this is wrong. His counter example is $X$ has a uniform distribution on $(0,1)$, with $S$ representing the event that $X \geq (1/2)$ and $R$ representing the event that $X > (1/2)$.

This shows the danger of my relying on intuition without formal training in Probability. Thus, the analysis below is invalid.

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Therefore:

$p(B|A) = 1 \implies [A \implies B] \implies [B^c \implies A^c] \implies $
$p\left(A^c|B^c\right) = 1.$

Answer 2

However, since I didn't at all use the Law of Total Probability as the question hinted at,

By Law of Total Probability: $\mathsf P((A\cup A^{\small\complement})\mid B^{\small\complement})=\mathsf P(A\mid B^{\small\complement}) +\mathsf P(A^{\small\complement}\mid B^{\small\complement})\\~\,\therefore~~~~\mathsf P(A^{\small\complement}\mid B^{\small\complement})=1-\mathsf P(A\mid B^{\small\complement})$

Which you used.