Okay. So B is base and R is rule. I get the whole idea now. Sorry about the other approach in the last question. We did recursive sets a different way.

So you know that 2 is in X and for every element x that is in X 10x and 4+x are also in X.

So 2 is in X because of B

20 is in X because of R1 (2 is in X so 10*(2) is in X)

6, 10, 14, 18, 22, 26 are all in X from R2 (2 is in X 2+4 = 6. 6+4 = 10 etc, so all are in X)

Also, since 20 is in X, 24 and 28 also are.

There are no odd numbers in X since 10 and 4 are even numbers so any combination of multiplying by 10 and adding 4 will still be even.

Or, more formally, if x is an even number, then 10x is even

If x is an even number, x+4 is even.

2 is even.

So all numbers in X are even.

So you know that 2 is in X and for every element x that is in X 10x and 4+x are also in X.

So 2 is in X because of B

20 is in X because of R1 (2 is in X so 10*(2) is in X)

6, 10, 14, 18, 22, 26 are all in X from R2 (2 is in X 2+4 = 6. 6+4 = 10 etc, so all are in X)

Also, since 20 is in X, 24 and 28 also are.

There are no odd numbers in X since 10 and 4 are even numbers so any combination of multiplying by 10 and adding 4 will still be even.

Or, more formally, if x is an even number, then 10x is even

If x is an even number, x+4 is even.

2 is even.

So all numbers in X are even.