Problem $6.(10$ points$)$ Consider the Ackermann function, a two$-$argument function where each argument takes a nonnegative integer value. It is defined as follows:

$($I$)$ If $m=0,$ then $A(m,n)=n+1.$

$($II$)$ If $m0$ but $n=0,$ then $A(m,n)=A(m-1,1).$

$($III$)$ If $m0$ and $n0,$ then $A(m,n)=A(m-1,A(m,n-1)).$

Compute $A(1,2)$ and $A(2,1).$ Show steps similar to the example on page $3$ of Lecture $7$ notes. You may use the value $A(1,1)$ computed there.