Term Test 1 Question 3

Hello fellows, today I'm going to tackle the question regarding structural induction. It is just something that I couldn't put my finger on at the first glance. I found this hard at the term test 1 so I'll take a few step down to look at the structural induction

So here's an explanation I found in the course notes:
Provided a recursive definition of a set, we naturally use structural induction to prove properties of its elements by using induction.
Suppose X is a defined set using induction and we want to prove that every element of X has a certain property P. A proof by structural induction proceeds in two steps.

Basis: We prove that every "smallest" or "simplest" element of X satisfy property P

Induction Step: We also prove that the larger elements that are made of these"smallest" or "simplest" elements satisfy property P

I didn't get the third question of term test 1 right so I'll take a few step to look into it.

Here's the question:
Define epsilon as the smallest set such that
1. Symbols x,y,z are elements of epsilon.
2. If expressions e1 and e2 are elements of epsilon, then so is (e1 + e2)
For e in epsilon, define vr(e) as the number of occurrences of symbols from the set {x,y,z} in e, and p(e) as the number o occurrences of parentheses from the set  {(,)} in e. For example, vr(((x+y)+z)) = 3 and p(((x+y)+z) = 4. Use Structural Induction to prove that for every e in epsilon, 2vr(e) = p(e) + 2

So as all induction proof, we start by defining the predicate: P(e): "2vr(e) = p(e) + 2"

First step would be looking for the basis

Basis: e in {x,y,z}: Each of these expressions has exactly one variable and 0 parentheses, so according to P(e) we have 2 * 1 = 0 + 2 = 2, so we know P(e) holds for e in the basis.

Secondly we show the induction step

Induction: Assume e1 and e2 are in epsilon and P(e1), P(e2)  [P(e1) and P(e2) have to hold in order to prove P((e1 + e2))

So 2vr((e1 + e2) =  2(vr(e1) + vr(e2))    #just rearrange it
                           =  2vr(e1) + 2vr(e2)
                           =  p(e1) + 2 + p(e2) + 2  #left hand side of the equation 2vr(e) = p(e) + 2
                           =  (p(e1) + p(e2) + 2) + 2
                           = p(e1 + e2) + 2  # because only two parentheses needed to wrap (e1 + e2)

So, for all e1 and e2 in epsilon, P(e1) and P(e2) implies P((e1 + e2)) by structural induction.