show $W_{t}^2 - t$ is an Ito process

Question

I am trying an example to show $W_{t}^2 - t$ is an Ito process where $W_t$ is a standard Brownian process

I am quite confused by the approach as the solution is simply using Ito's lemma to write out:

$$df(W_t,t) = 2W_{t}dW_{t} \tag1$$

$$f(W_t,t) -f(W_0,0) = W_{t}^2 - t = \int_{0}^{t} 2W_{s}dW_{s} \tag2$$

First, I am not quite sure why this is a prove as it is simply applying the Ito's lemma? As normally, in my mind, if I want to prove A is B, then I would normally check the characteristics of A against criteria to be B but in this question, I don't feel like the solution is doing this.

Secondly, in (2), how does the integration result in $ W_{t}^2 - t$? where does the $t$ come from?

Answer 1

They are defining $f(u, v) := u^2 - v$ so that $f(W_t, t)=W_t^2-t$ and $f(W_0, 0) = W_0^2 - 0 = 0$. (1) is the result of applying Itô's lemma to this choice of $f$, and (2) is rewriting (1) as an integral instead of as differentials.

For why this shows $W_t^2-t$ is an Itô process, check the definition of an Itô process.