Bayesian Inversion Review I am looking over a paper currently and am stuck on a particular step seeing how the intuition works. I know this is quite elementary but it is known:
$$P(ab|c)=P(a|c)P(b|c)$$
This being said I am confused how
$${P(a|c,d)\over{P(b|c,d)}}={P(c|a,d)\over{P(c|b,d)}}{P(a|d)\over{P(b|d)}}$$
If possible I would like to see the intuition in with how the above Bayesian inversion is done.
Thanks. 
P.S. I also posted this on the Math Stackexchange site get help as well so if this is inappropriate please let me know so that I'll remove this question
 A: Ami Tavory provides a nice proof of the equation. Indeed you can show equality by multipling both sides of the equation in Bayes theorem, $p(x \vert y) = p(y \vert x) \tfrac{p(x)}{p(y)}$ by $p(y)$, and using the rule $p(a \vert b) \cdot p(b) = p(a,b)$. His answer is a variation of that proof for Bayes theorem. 
Here is a slightly different path. 
I alway liked the quadrant structure as explained on wikipedia:
https://en.wikipedia.org/wiki/File:Bayes_theorem_visualisation.svg
In that expression you only have to add this extra parameter.
$\begin{array}
\\
P(a,c \vert d) &= P(a \vert c,d) \cdot P(c \vert d)\\
 &= P(c \vert a,d) \cdot P(a \vert d)
\end{array}$
analogous to the rule $P(a,c) = P(a \vert c) P(c) $ but just with the extra condition
and from equating both sides a version of Bayes theorem will follow
$P(a \vert c,d) = P(c \vert a,d) \frac{P(a \vert d)}{P(c \vert d)} $
doing the same for $P(b \vert c,d)$, taking the ratio, and eliminating $P(c \vert d)$ provides the given expression
$\frac{P(a \vert c,d)}{P(b \vert c,d)} = \frac{P(c \vert a,d) \frac{P(a \vert d)}{P(c \vert d)}}{P(c \vert b,d) \frac{P(b \vert d)}{P(c \vert d)}} = \frac{P(c \vert a,d) P(a \vert d)}{P(c \vert b,d) P(b \vert d)}$
My intuitive view is always in terms of a likelihood function, which we often use in statistics. 
$p_{post} = \mathcal{L}\cdot p_{prior}$
A: Say you take the RHS of the equation, and multiply the numerator and denominator by $P(d)$:
\begin{equation}
\frac{P(c|a, d)}{P(c|b, d)}
\frac{P(a | d)}{P(b | d)}
=
\frac{P(c|a, d)}{P(c|b, d)} 
\frac{P(a | d)}{P(b | d)}
\frac{P(d)}{P(d)}
=
\frac{P(c|a, d)}{P(c|b, d)} 
\frac{P(a, d)}{P(b, d)} 
=
\frac{P(c, a, d)}{P(c, b, d)} 
\end{equation}
Multiplying the numerator and denominator of the LHS by $P(c, d)$, you'd get the exact same fraction.
