Are cards 12 and 15 redundant?

[jatubio]

I just played this game: https://boardgamearena.com/gamereview?table=777530059

And I discarded the option 'Blue=Purple' on card D (12), because I thought that if Blue=Purple, then the 'Yellow>Blue and Purple' will be the only choice posible on card E (15) making it redundant.

But the solution was: 252. So I have the doubt why 'Blue=Purple' on card 12 doesn't make card 15 redundant?

[Jellby]

That Y>B,P is the only choice doesn't make the card redundant. Without it, you could have Y=B,P or Y<B,P.

Given the other cards:
B < 3
Y odd
no 3s
B = P
even > odd

We can deduce the code is 212 or 252. Card 15 makes it unique, so it's not redundant.

[Romain672]

With just ABC, you got:
B=12, Y=15, P=1245
If you add F (more evens), you know Y is odd, so both B and P are even. Which mean:
B=2, Y=15, P=24.

Then if you add only D (B=P) (and 'remove E'), you have 212 and 252 as solutions.
If you add only E (Y>BP) (and 'remove D'), you have 252 and 254 as solutions.

So none of them are redundant, and if you add both of them you are left with only 252.

I discarded the option 'Blue=Purple' on card D (12), because I thought that if Blue=Purple, then the 'Yellow>Blue and Purple' will be the only choice posible on card E (15)

This is a wrong reasonning.
When looking at the start of the game on D&E, you know that Y>BP else D would be redundant in others cases.
But even if with B=P you know that E said that Y>BP, that doesn't mean that E is redundant/doesn't give new information: E will still give you the information that Y is the highest.

[jatubio]

Ohhhhh! Thank you both very much for your answers and explanations!! It seems like I have misinterpreted the "redundant rule"!