x is bound, y is free; the formula is therefore not
closed.
true in assignment [y |-> 0] over the domain of
natural numbers with the usual interpretation of
` <= `, because 0 <= x is true for every natural number
x:
0 <= 0, 0 <= 1, 0 <= 2, ...
(existsx: x | 15)
x is bound; the formula is closed.
true for every assignment over the natural numbers with `|'
interpreted as `divides', because x | 15 is true for some natural
number x (e.g. for 3):