Identity: $P(A \mid B) > P(A \mid B^c)$ is equivalent to $P(A \mid B) > P(A)$

I can show that this identity is true using algebra.

We know if $$P(A \mid B) > P(A \mid B^c),$$ then $$P(A\cap B) > P(A)P(B).$$

However, I am trying to understand my textbook's arugment using the law of total probabilty.

It states that since $$P(A) = P(A \mid B)P(B) + P(A \mid B^c)P(B^c),$$

then $P(A \mid B) > P(A \mid B^c)$ must imply that $P(A \mid B) > P(A)$.

I am lost.

Also this is my first question I am asking on here. Open to criticism on how I could ask better questions.

Thanks.

$$P(A) = P(A \mid B) P(B) + \color{blue}{P(A \mid B^c)} P(B^c) < P(A \mid B) P(B) + \color{blue}{P(A \mid B)} P(B^c) = P(A \mid B).$$