Show for closed terms $t$ and formulas $\phi$. Given a structure $M$: $M\vDash t = \overline{t}\,^M$

Question

Show for closed terms $t$ and formulas $\phi$. Given a structure $M$: $$ \begin{align} M&\vDash t = \overline{t}\,^M\\ M&\vDash \phi(t) \leftrightarrow \phi(\overline{t}\,^M) \end{align} $$

For the first statement, we can say that any interpretation of a closed term is the same term, so the valuation of this atomic formula gives always $1$. The other part is not than clear, because we see a formula, not necessary a sentence, so, the valuation ($1- |(\phi(t))^M -\phi(\overline{t}\,^M)|$) must be $1$. I' not sure what is the next step. How do you prove that?

Answer 1

Comment

we can say that any interpretation of a closed term is the same term;

not exactly.

If $t$ is a (closed) term of the language, then $t^{\mathcal M}$ is its interpretation in the structure $\mathcal M$ while $\overline {(t^{\mathcal M})}$ is the "added" constant symbol for $t^{\mathcal M} \in |\mathcal M|$ [see page 65].

Example from the language for f-o arithmetic [see Exercise 1, page 68], i.e. for the structure :

$\mathcal N = \langle \mathbb N ,+,·,S,0 \rangle$.

A closed term is e.g. :

$S(0)+S(0)$;

if we call it $t$, then $t^{\mathcal N}$ is the number $2$, while the "added" constant symbol $\overline {(t^{\mathcal N})}$ is $\overline 2$.

Thus, applying clause 2 of Definition 3.4.2, we have to prove that:

$[t = \overline {(t^{\mathcal M})}]_{\mathcal M}=1$

and this is so if:

$t^{\mathcal M} = \overline {(t^{\mathcal M})}^{\mathcal M}$

which is "convoluted" but obvious: the "added" constant symbol $\overline {(t^{\mathcal M})}$ is the "name" for the object associated in the interpretation to the term $t$. Thus $\overline {(t^{\mathcal M})}^{\mathcal M}$ is the object itself.

But the object associated in the interpretation to the term $t$ is $t^{\mathcal M}$.

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Regarding the formula $\varphi(t)$, it is closed and thus a sentence; so, we can apply Lemma 3.4.5 (v).