First, let's recap how mathematical induction works. To prove a statement $\Phi(n, \vec{w})$, where $n \in \mathbb{N}$ is the induction variable and $\vec{w}$ are any additional parameters, we need two things:

- base step: $\forall \vec{w}: \Phi(1, \vec{w})$ holds.
- inductive step: $\forall \vec{w} \; \forall n \in \mathbb{N} : \Phi(n, \vec{w}) \rightarrow \Phi(n + 1, \vec{w})$ holds.

Now, let's have a look at the inductive step in the proof:

$$
\forall a \in \mathbb{N} \; \forall b \in \mathbb{N} \; \forall n \in \mathbb{N} [(\max(a - 1,b - 1) = n \rightarrow a - 1 = b - 1) \rightarrow (\max(a, b) = n + 1 \rightarrow a = b)]
$$

This is actually a true statement, but there are pathological cases when $a -1 < 1$ and $b - 1 < 1$. However, we can easily fix it by adding $1$:

$$
\forall a \in \mathbb{N} \; \forall b \in \mathbb{N} \; \forall n \in \mathbb{N} [(\max(a ,b) = n \rightarrow a = b) \rightarrow (\max(a + 1, b + 1) = n + 1 \rightarrow a + 1 = b + 1)]
$$

Yet again, this is a true statement and without any pathological cases. We can validate the base case and surely enough, it is true. Does it mean the proof is correct after all?

We have one more hurdle to clear. We need to check that the inductive step has the correct form mandated by mathematical induction. Namely, let's extract $\Phi(n, \vec{w})$. One option is:

$$
\Phi(n, \vec{w}) \equiv \Phi(n, a, b) \equiv \max(a ,b) = n \rightarrow a = b
$$

Let's plug that back into the inductive step prescription and check that we get the exact same inductive step as was proposed:

$$
\forall a \in \mathbb{N} \forall b \in \mathbb{N} \forall n \in \mathbb{N} [(\max(a,b) = n \rightarrow a = b) \rightarrow (\max(a, b) = n + 1 \rightarrow a = b)]
$$

But wait, that's not quite the same thing as was proposed! Maybe we've just used the wrong formula, let's try the other option:

$$
\Phi(n, \vec{w}) \equiv \Phi(n, a, b) \equiv \max(a + 1 ,b + 1) = n + 1 \rightarrow a + 1 = b + 1
$$

Oh no, that's not working either:

$$
\forall a \in \mathbb{N} \forall b \in \mathbb{N} \forall n \in \mathbb{N} [(\max(a + 1,b + 1) = n + 1 \rightarrow a = b) \rightarrow \\ (\max(a + 1, b + 1) = n + 2 \rightarrow a + 1 = b + 1)]
$$

So the problem here is that the original proof is not using mathematical induction correctly and hence we can rule it out.

However, let's see if the theorem still holds with the corrected inductive step. And it is easily seen that with the right combination of parameters, it does not. If we put $n=1$, $a=1$, $b=2$, this is true:

$$
max(1,2) = 1 \rightarrow 1=2
$$

but this is false:

$$
max(1,2) = 2 \rightarrow 1=2
$$

This rules out the corrected inductive step as well.