[∃ B: P(A|B) ∈ (0,1)] → [P(A) ∈ (0,1)]. Better?

Yes, it makes it clearer what you're doing wrong. I'll do what I should have done earlier, and formalize my argument:

Let's call "2+2=4" A, "2+2=3" B, "I can visualize 2+2=4" C, "I can visualize 2+2=3" D, "I can remember visualizing 2+2=4" E, "I can remember visualizing 2+2=3" F.

So, my claim is that P(A|C) is 1, likewise P(B|D). (Remember, I don't think it's like this in real life, I'm trying to show that the argument put forward to prove that is not sufficient.)

What is the Bayes formula for tomorrow's assessment?

Not, P(A|C,D), which (if <1) would indeed disprove P(A|C)=1.

But, instead, P(A|E,D). This can be less than 1 while P(A|C)=1. I'll just make up some arbitrary numbers as priors to show that.

I'm assuming A and B are mutually exclusive, as are C and D.

P(A)=.75

P(B)=.25 (just assume that it's either 2 or 3)

P(C)=.375

P(D)=.125

P(memory of X | X happened yesterday)=.95

P(memory of X | X didn't happen yesterday)=.001

P(E)=P(C)*.95+P(~C)*.001=0.356875

P(F)= P(D)*.95+P(~D)*.001=0.119625

P(C|A)=.50

P(C|B)=0

P(D|A)=0

P(D|B)=.50

P(A|C) = P(C|A)P(A)/P(C)=(.50*.75)/(.75*.50+.25*0)=1

P(A|C,D) is undefined, because C and D are mutually exclusive (which corresponds to not being able to visualize both 2+2=3 and 2+2=4 at the same time)

P(F,D)=P(D)*.95=0.11875

P(A|E,D)= P(E,D|A)P(A)/P(E,D)=0 (Because D|A is zero).

Using my numbers, you need to derive a mathematical contradiction if there are, truly "technical reasons" for this being impossible.

The mistake you (and EY) are making is that you're not comparing P(A) to P(A|B) for some A,B, but P(A|B) to P(A|C) for some A,B,C.

Added: I made two minor errors in definitions that have been corrected. E and F are not exclusive, and C and D shouldn't be defined as "current", but rather as having happened, which can only be confirmed definately if they are current. However, they have the evidential power whenever they happened, it's just if they didn't happen now, they're devalued because of fragile memory.

Added: Fixed numerical error and F where it was supposed to be E. (And errors with evaluating E and F. I really should not have assumed any values that I could have calculated from values I already assumed. I have less degrees of freedom than I thought.)