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a@b = a+(b+1)
Since ordinary addition is associative and commutative,
we may write
a@b = a+(b+1) = a+b+1
The rule for @ is "add and then add 1"
We must prove that
(a @ b) @ c = a @ (b @ c)
We start with the left side and end up with the right side:
(a @ b) @ c
Substitute a+b+1 for a @ b
(a+b+1) @ c
Add them and then add 1
(a+b+1)+c+1
a+b+1+c+1
a+b+c+2
a+b+c+1+1
Put in grouping symbols:
a + (b+c+1)+1
This is adding a and (b+c+1) and then adding 1, so that is:
a @ (b+c+1)
In the parentheses is adding b and c and then adding 1,
so that is
a @ (b @ c)
So therefore
(a @ b) @ c = a @ (b @ c)
Edwin