Twelve for eight

I had twelve cards and my friend had eight. Two places in my album were empty, and the two rare cards that fitted them were in my friend's pile.

I had $12$ cards, my friend $8$. My album had two gaps, and the two rare cards were his.

Cards: me $m=12$, friend $f=8$. I need $2$ rare cards $r$.

In our game one rare card is worth three common ones. So two rare cards are worth six common ones, and I offered six of my doubles.

A rare card $r$ and a common card $c$ obey $r=3c$, so $2r=6c$: I gave $6$ doubles.

Common card $c$: $r=3c$, so $2r=6c$.

My friend agreed. After the swap I had eight cards and he had twelve, which looked like a bad deal for me.

After the swap I had $12-6+2=8$ cards and he had $8+6-2=12$.

Then $m^{\prime }=12-6+2=8$ and $f^{\prime }=8+6-2=12$.

But the total was still twenty, nothing was lost, and my album was full at last. A swap makes no new cards; it moves each card to the hands where it is worth more.

The total is $8+12=20=12+8$. Nothing was lost, and my album is full.

$m^{\prime }+f^{\prime }=m+f=20$ and $m^{\prime }<f^{\prime }$, yet the album is full.