How do I use structural induction to show that for all $(a,b) in S$ that $(a+b) = 4k$ for some $k in Bbb Z$?

Let $p$ and $q$ denote, respectively, the operation $$(a,b)mapsto (a+1,b+3)$$ and the operation $$(a,b)mapsto (a+3,b+1)$$ for each $(a,b)in S$. For each pair $(a,b)in S$, let $mu(a,b)$ denotes the minimum number of times the operations $p$ and $q$ are required to reach $(a,b)$, starting from $(0,0)$. We claim that $$a+b=4,mu(a,b),.$$

We shall induct on $mu(a,b)$. If $mu(a,b)=0$, then $(a,b)=(0,0)$. Clearly, $$a+b=0=4cdot 0=4,mu(a,b),.$$ From now on, we suppose that $mu(a,b)>0$. Hence, in a minimum sequence of operations to get $(a,b)$ from $(0,0)$, $(a,b)$ can be obtained from some $(a',b')in S$ by either a use of $p$ or a use of $q$.

If $(a,b)$ is obtained from $(a',b')$ via the use of $p$, then $$(a,b)=(a'+1,b'+3),.$$ Therefoore, $$a+b=(a'+1)+(b'+3)=(a'+b')+4,.$$ Using the induction hypothesis, $a'+b'=4,mu(a',b')$. Thus, $$a+b=4,mu(a',b')+4=4,big(mu(a',b')+1big),.$$ Obviously, $mu(a,b)=mu(a',b')+1$. Therefore, $a+b=4,mu(a,b)$, as required.

If $(a,b)$ is obtained from $(a',b')$ via the use of $a$, then $$(a,b)=(a'+3,b'+1),.$$ Therefoore, $$a+b=(a'+3)+(b'+1)=(a'+b')+4,.$$ Using the induction hypothesis, $a'+b'=4,mu(a',b')$. Thus, $$a+b=4,mu(a',b')+4=4,big(mu(a',b')+1big),.$$ Obviously, $mu(a,b)=mu(a',b')+1$. Therefore, $a+b=4,mu(a,b)$, as required.

Remark. In fact, $mu(a,b)$ is the number of times the operations $p$ and $q$ are required to reach $(a,b)$ from $(0,0)$, not just the minimum number. Furthermore, it can be shown that all $(a,b)in S$ such that $mu(a,b)=m$ for a given $minmathbb{Z}_{geq 0}$ are of the form $$(m,3m),(m+2,3m-2),(m+4,3m-4),ldots,(3m,m),.$$ For $k=0,1,2,ldots,m$, the element $(m+2k,3m-2k) in S$ requires (in any order) $m-k$ times of the operation $p$ and $k$ times of the operation $q$. That is, $$S=big{(0,0),(1,3),(3,1),(2,6),(4,4),(6,2),(3,9),(5,7),(7,5),(9,3),ldotsbig},.$$