The symbol $\sigma[x/a]$ in mathematical logic

Question

I've just started learning mathematical logic, and my teacher teaches in a very formal and unintuitive way.

I'd like to know what meaning hides behind the symbols below:

For an assignment $\sigma: Terms \rightarrow|M|$ (where M is a structure) we define:

• $\sigma[x/a](x) = a$
• $\sigma[x/a](y) = \sigma(y)$ (for x$\neq$ y)

Does this mean that $\sigma$ just replaces $x$ with $a$, and we don't know what else it does?

And another related thing: When I'm trying to formally prove $"(\forall x)(\exists x)(x < x)"$ is always False, I end up with $\left(\sigma[x/a]\right)[x/b](x)$ (with negations and unions). What's the meaning of this expression?

Answer 1

No, intuitively $\sigma[x/a](x)$ is the function that's exactly the same as $\sigma$ except it maps $x$ to $a$ instead of whatever $\sigma$ maps $x$ to.

Answer 2

We have to "calculate" the truth value of :

$(∀x)(∃x)(x < x)$

using the truth conditions :

$[∀x\varphi](\sigma) = 1$, if $[\varphi](\sigma[x/a]) = 1$, for all $a$

and

$[∃x\varphi](\sigma) = 1$, if $[\varphi](\sigma[x/a]) = 1$, for some $a$.

Assume as domain $\mathbb N$; clearly : $n < n$ is false for every $n \in \mathbb N$.

Thus we have $[(∃x)(x < x)](\sigma) = 0$, because $[(x < x)](\sigma[x/a]) = 0$ for every $a$.

But $x$ is not free in $(∃x)(x < x)$; thus, for $b$ whatever, $[(∃x)(x < x)]((\sigma[x/a])[x/b]) = 0$.

But $(\sigma[x/a])[x/b] = \sigma[x/b]$, and thus :

$[(∀x)(∃x)(x < x)](\sigma) = 0$.