x-(y-z) = (x-y)+z

How to prove that x-(y-z) = (x-y)+z ?

We have at our disposal:
-Axioms determining Natural numbers
-[a,b]R[c,d] is true when a + d = b + c (This has something to do with defining the axioms for integers, but how to apply it to this situation?)

If anybody asks, it's N.L. Biggs 7.5.3

Thank you in advance

Can you not just set
y-z = a
fiddle to get -a = z-y
Then x-(y-z)= x-a = x+ (z-y)
= (x-y)+z  using associtativity and commutivitiy

>>1
Can't be done. You're black.
>>2
Is so black, he makes Wesley Snipes look pale.

x - (y - z)
x + (-1) * (y + (-1) * z) (equivalent from definition of negative numbers)
x + (-y + 1 * z) (distributivity of multiplication)
(x + (-y)) + z (associativity of addition)
(x - y) + z

This is the sort of thing they teach in elementary school.