Or less formally, it's proof that if a test has 50% false positive results and 50% false negative results then probability of tested hypothesis will stay the same even after the test.
It seems to be obvious, but I want to be sure, so here goes the proof
Suppose we have a test to tell sick people apart from healthy people. The test is far from prefect, it has 50% false positives and 50% false negatives. Even worse, we suspect that the test is absolutely useless to help us to detect sickness, that even after taking into account its evidence we won't be more or less sure about sickness of the tested person. Or more formally, we suspect that after taking the test following thing will always be true: P(H|e)=P(H). From now on "H" means event of the tested person being sick, while "e" means event of getting the positive test result.
Let's check. We know that the test has 50% false positives and 50% false negatives, meaning that we have P(e|~H)=1/2 and P(~e|H)=1/2 respectively. Also let's notice that there are only 4 possible combinations of events, namely H and ~e, H and e, ~H and e, ~H and ~e. Let's name numbers of such combined events as a,b,c, and d respectively (for example, "a" corresponds to number of times combined event of H and ~e happened).
Here is a picture for better understaning:
It means that P(~e|H) can be written as a/(a+b) and P(e|~H) is equal to c/(c+d). Let's concentrate on a/(a+b). Due to transitivity a/(a+b)=1/2, also implying that "a" can't be equal to 0. If we flip both equal fractions upside down the equality will be preserved, thus we get (a+b)/a=2. "(a+b)/a" can be expressed as sum of two fractions, "(a/a) + (b/a)", thus we get 1+(b/a)=2. From this we simplify to b/a=1. Equality b/a=1 implies that a=b. Similiarly we can prove that c=d.
Let's remember that P(H|e)=P(H). It can be expressed as b/(b+c)=(a+b)/(a+b+c+d). If we remember that a=b and c=d, the equality can be simplified to a/(a+c)=a/(a+c) and it becomes obvious that this equality will always hold (except when a+c=0). And we are done, we successfully deduced that truth of P(e|~H)=0.5 and P(~e|H)=0.5 will always lead to P(H|e)=P(H) being true.
P.S. By using the shortcut that drhab has shown, I was also able to prove that if P(e|~H)=0.5 and P(~e|H)=0.5, then P(H|~e)=P(H). In other words, it doesn't matter if the test result is negative or positve, both kinds of evidence will be neutral (i.e. they won't affect probability of hypothesis H)
P.P.S By using method similar to the one I employed (i.e. NON-shortcut method, the longer path) I was also able to prove that if P(e|~H)=P(~e|H)=0.5 , then P(H|e)=P(H|~e)=P(H). I applied it two times, the first time for P(H|e)=P(H) and the second time for P(H|~e)=P(H)
Correct.
Your way to find $a=b$ on base of $\frac{a}{a+b}=\frac12$ is quite cumbersome though. Just multiply both sides with $(a+b)2$ and then subtract $a$ on both sides.
Shortcut:
Note that $P(e\mid H)=1-P(\sim e\mid H)=1-\frac12=\frac12$.
Then the rule of Bayes tells us that:$$P(H\mid e)=\frac{P(e\mid H)P(H)}{P(e\mid H)P(H)+P(e\mid\sim H)P(\sim H)}=\frac{\frac12P(H)}{\frac12P(H)+\frac12P(\sim H)}=P(H)$$