Confused by a step in the 'Rule C' proof in Mendelson's Logic Textbook

Question

I've been working through 'Introduction to Mathematical Logic, 5th Ed' by Mendelson, and I've found a step in the proof of proposition 2.10 (Rule C) that I cannot understand.

In the proof of this proposition, on what is to me the fifth line, Mendelson writes:

'We replace $d_k$ everywhere by a variable $z$ that does not occur in the proof.'

I do not understand how Mendelson is justified in doing this step, it doesn't seem valid.

Answer 1

The step is valid.

Simplifying a little bit the case, we have to show that:

if $\Gamma \vDash \mathcal C(d) \to \mathcal B$, then $\Gamma \vDash \mathcal C(z) \to \mathcal B$,

provided that we have proved $\exists x \mathcal \ C(x)$.

Assume not, i.e. that we have an interpretation $M$ with domain $D$ and a sequence $s$ such that :

$M \vDash \Gamma, M,s \vDash \mathcal C(z)$ and $M \nvDash \mathcal B$.

But we have proved $\exists x \mathcal \ C(x)$ and this means that, in every interpretation $M$: $M \vDash \exists x \mathcal \ C(x)$.

This means that there is a sequence $s'$ such that $s'^*(x) \in D$ that satisfies $\mathcal C(x)$.

By Rule C, we have assigned to this element the "name" $d$, where $d$ is the new individual constant introduce with the rule.

Thus, we have that:

$M \vDash \Gamma, M \vDash \mathcal C(d)$ and $M \nvDash \mathcal B$

contradicting the fact that: $\Gamma \vDash \mathcal C(d) \to \mathcal B$.