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Yes, because suppose that 'a' and 'b' are both arbitrary integers. Then (a-b) or (b-a) will then provide you with another integer. Suppose that the integer you are given from (a-b) is not unique. Then we have:

(a-b)=c

and

(a-b)=c'

Then, trivially, since (a-b)=(a-b), we have c=c'.

Thus it is closed under subtraction.

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