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