**isSubmapOf**

*O(n+m)*. The expression (isSubmapOfBy f m1 m2) returns True if all keys in m1 are in m2, and when f returns True when applied to their respective values. For example, the following expressions are all True:
> isSubmapOfBy (==) (fromList [(1,1)]) (fromList [(1,1),(2,2)])
> isSubmapOfBy (<=) (fromList [(1,1)]) (fromList [(1,1),(2,2)])
> isSubmapOfBy (==) (fromList [(1,1),(2,2)]) (fromList [(1,1),(2,2)])
But the following are all False:
> isSubmapOfBy (==) (fromList [(1,2)]) (fromList [(1,1),(2,2)])
> isSubmapOfBy (<) (fromList [(1,1)]) (fromList [(1,1),(2,2)])
> isSubmapOfBy (==) (fromList [(1,1),(2,2)]) (fromList [(1,1)])
*O(n+m)*. The expression (isSubmapOfBy f t1 t2) returns True if all keys in t1 are in tree t2, and when f returns True when applied to their respective values. For example, the following expressions are all True:
> isSubmapOfBy (==) (fromList [('a',1)]) (fromList [('a',1),('b',2)])
> isSubmapOfBy (<=) (fromList [('a',1)]) (fromList [('a',1),('b',2)])
> isSubmapOfBy (==) (fromList [('a',1),('b',2)]) (fromList [('a',1),('b',2)])
But the following are all False:
> isSubmapOfBy (==) (fromList [('a',2)]) (fromList [('a',1),('b',2)])
> isSubmapOfBy (<) (fromList [('a',1)]) (fromList [('a',1),('b',2)])
> isSubmapOfBy (==) (fromList [('a',1),('b',2)]) (fromList [('a',1)])