> "insert into TABLE_B (ID, TABLE_A_ID) values (1, 5);"
> Why does the second transaction wait?

I might be thinking too simple, but for me this looks obvious.

If the first transaction translates to "Update Table_A set ID = 5
where ID = 5", it does not make too much sense, but should not affect
the "child" table. In this case, I suppose that Firebird server is not
prepared for nonsense commands but expects all updates to be real
updates (changing indeed the ID to another value).

But any attempted change of the ID to another value *must* stop
the second transaction, because otherwise it would be next to
impossible to keep database consistency. The foreign key relation
would be corrupted!

So, I guess what you are asking is: why is the parser not intelligent
enough to ignore my attempt to update the ID with the same value?

If I had a vote on this, I would like to keep it as it is.
Otherwise the parser would have to check this on every update which
surely is some overhead; to be paid by each and every update
transaction.

ciao,
André