Inverted Minors with a big fit for partner's minor

A couple of hands have come up recently that have illustrated a pet theory of mine regarding inverted minors. They haven't happened to me personally but have been related to me. Take this one, for instance:

♠KQ92 ♥A6 ♦2 ♣AJT952. Matchpoints, favorable vulnerability. LHO passes as dealer and partner opens 1♣!

Now, it's a well-known principle of 2/1 bidding* that when we have a hand good enough to force to game, we show our longest suit first. Guess what? We have enough to force to game and our longest suit happens to be clubs.

So, let's start with 2♣, for now ignoring the spade suit. Partner probably has a balanced hand with three or maybe four clubs but that's not important for now. Our job is to bid our hand. Of course, if partner has only three clubs, then his automatic rebid is 2NT.  Anything else would be uncivilized. No, seriously, Howard Piltch taught me many years ago that you may not bid a new suit as rebid if you don't possess four of your minor. I still agree with that scheme. And, furthermore, 2NT is forcing! The only non-forcing bid is 3m.

So, assuming four clubs, partner's rebid over 2♣ is likely to be 2♥ or 2♠ (or possibly 2NT with stoppers in both suits). If partner rebids 2♥, he most probably doesn't have four spades (although he could be 4414 or 4405). Our 2♠ will simply be taken as a fragment and no harm is likely to be done.

But, it should be allowable for responder to raise 2♥ or for opener to raise 2♠ to show four. After all, most of the time these raises never occur because the emphasis is always on finding stoppers for no-trump. But if they do happen, the meaning should be clear: a potential double fit has been found. Since responder would never raise the minor without a game-forcing hand with an eye towards 6m (clubs in this case), it's hard to imagine anything going seriously wrong, provided that both partners are open to this possibility.

In the case cited above, the hand opposite was: ♠A74 ♥KJT9 ♦JT3 ♣K86. The club slam is cold** and much of the time, will make seven unless LHO happened to overcall 2♣ with 2♦. But, in practice, almost every table played the hand in 3NT which not only goes down on a diamond lead, but obviously wouldn't score anywhere near as well as 6♣.

So, here's my suggested bidding sequence: (p) 1♣ (p) 2♣ (2♦) p (p) 2♠ p 2NT (p) 4♦ (or other keycard ask, followed by the two-key-card response...) 6♣.

If you start with 1♠, it will be almost impossible to persuade opener that you have six clubs in addition to four spades, which is why nobody got to 6♣. I wouldn't advocate suppressing the major with, say, a five-card minor and a relatively balanced hand. That would be taking things too far. Yet I expect most people will disagree with doing it even in this situation because it was fed to them with their mother's milk that you may not suppress a four-card major when you raise partner's opening minor. But there's a time to break just about every rule. And this is one of those times.

*  inverted minors are often used in "standard" but they were invented for a 2/1 game-forcing context so that's what I'm assuming here.

** unless you decide on a backward finesse against the club queen, you will see her pop up before you have to decide finesse-or-drop.