Hey, I just played a game where this problem accured.

I am German and have the German cards/games.

I checked the rules of the Alchemy set.

The German ruleset says:

$2P > $2 – This is incontestable.

It also states that $2p is not more than $3 ($2P <= $3 (?) because not more =/= less (?))

It should be reasonable that one can take a $2 card with Haggler if buying a $2P card.

Since $2P <= $3 (?) and <= or < should be transitive, it should follow:

$2P <= $3 < $4 --> $2P < $4 (if < and <= are transitive!).

That means it is possible to obtain cards with Potion costs using Haggler even when the card bought didn’t cost any Potions.

Open problem: could one take a $2P card after buying a $3 card (does $2P < $3 hold?)?

Because the (at least German) rules give upper and lower bounds for cards with potion costs, I think it should be reasonable that all costs can be ordered via a total order. And since every relation (except for the problem case mentioned above) seems to be strict I would argue further that the relation of $2P and $3 should be $2P < $3. But that is interpretative in my opinion.

That would leave a strict total order as follows: $a + bP < $c + dP iff (a < c) or (a = c and b < d).

So one could simply say that a Potion in cost can be interpreted as the card’s cost + 0.5 if a card only speaks of the cost of a card. (This should’t, of course, be true if a card explicitly just talks about the $-costs, e.g., take a card for $5, since the rules say so.)

A completely different approach is shown in the following figure:

This could be true if ¬($2P > $3) is not equal to $2P <= $3.

This means that the order is not total but partial.