Bayesian probability: proving $P(x|y,x) = 1$

Question

I have an exercise where I'm supposed to prove that $P(x|y,x) = 1$

I've come up with the following but I'm not sure if it feels right to me:

$P(x|y,x) =$

$P(x,y,x) / P(y,x) = $

$P(x|y,x)P(y|x)P(x) / P(y,x) = $

$P(x|y,x)P(y|x)P(x) / P(y|x)P(x) = $

We now have

$P(x|y,x) = P(x|y,x)$

So:

$P(x|y,x) / P(x|y,x) = P(x|y,x) / P(x|y,x) = 1$

Is this correct?

Answer 1

Assuming that $P(x|y,x) = P(x|y \cap x)$, just note that $$x \cap y \cap x = y \cap x \Rightarrow P(x|y \cap x) = \frac{P(x \cap y \cap x)}{P(y \cap x)} = \frac{P(y \cap x)}{P(y \cap x)} = 1$$