Prove that for any odd number m, there is some $e \in \mathcal{E}$ such that $vr(e) + op(e) = m$
vr means variable, op means operators like $+/-$
Attempt:
I will prove this with structural induction
Base Case: $3$ cases: $e = x, y, z$
let $e = x$, then $vr(e) = 1$ and $op(e) = 0, 1 = 0 + 1 = 1$. as wanted
Similarly for $e = y, z$
Inductive step: let $e_1, e_2 \in \mathcal{E}$. Let $n$ be an odd number. Suppose $n = vr(e_1) + op(e_1)$ and $n = vr(e_2) + op(e_2)$
We need to show $(e_1 + e_2), (e_1 - e_2), (e_1 \div e_2), (e_1 \times e_2) \in \mathcal{E}$
Note:
$vr(e) = vr(e_1) + vr(e_2)$
$op(e) = op(e_1) + op(e_2) + 1$
let $e = e_1 + e_2$
$n + 2 = vr(e) + op(e)$ [By definition]
$= vr(e_1) + vr(e_2) + op(e_1) + op(e_2) + 1$ [I.H]
$= n + n' + 1$ [I.H]
$= 2n + 1$
Not sure what I'm doing after inductive step.