The situation is now

AKx
J
Q10x
--

xxx
x
--
10xx

If you have played a spade winner, nothing is changed. If you have played your heart winner, the squeeze occurs on the last club. I will describe it as is.

There are simple squeeze possibilities if spades are 5-2, but the squeeze I will describe works even when spades are 4-3. It requires Righty to hold the king of diamonds, which seems likely given that Lefty did not lead diamonds on opening lead and led his ace when he did lead diamonds. It also requires Lefty to have the jack of diamonds. There is no reason to place him with this card; this is just the best chance for making the rest of the tricks. (Most of your other chances also require Lefty to have the jack of diamonds.)

First, play out all of your club winners except for one. You pitch a diamond and a spade from the board. That leaves this position:

AK
J
Q10
---

xxx
x
---
x

So far, no one is squeezed. Assuming no one has pitched any useful cards, the situation around the table is

            AK
            J
            Q10
            ---
Jxx                  Qxx
---                  ---
Jx                   Kx
---                  ---
            xxx
            x
            ---
            x

But the small heart to the jack squeezes both opponents. If Lefty pitches a diamond, you can lead the queen of diamonds, smothering the jack. So suppose Lefty pitches a spade. If Righty pitches a diamond, you can lead a small diamond and ruff out the king. So suppose Righty pitches a spade. Now your long spade is good -- you can cash the AK of spades and come to your hand with a trump ruff.

I know of no standardized terminology for this squeeze. I call it a two-suited double trump squeeze. It works because both opponents must save their diamonds, so both are squeezed out of their spades. (Normally, if you just have only two threats, there is no double squeeze, because one opponent can guard against one threat and the other opponent can guard against the other threat.)

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