How to understand one of the properties of satisfaction of wfs in first-order logic?

Question

Proposition:

If the free variables (if any) of a well-formed formula $\textbf{B}$ occur in the list $x_{i_1}, \dots, x_{i_k}$ and if the sequences $s$ and $s^′$ have the same components in the ${i_{1}}^\text{th}, \dots, > {i_{k}}^{th}$ places, then $s$ satisfies $\textbf{B}$ if and only if $s^′$ satisfies $\textbf{B}$. Here $s$ and $s^′$ are denumerable sequences $s = (s_1, s_2, \dots )$ of elements from the domain $D$ of some interpretation $M$.

It is from Mendelson's book "Introduction to Mathematical Logic". The problem here is that I can't fully agree with the proposition, hence can't approach to prove it as it doesn't make sense to me (although obviously it is just me missing something).

My understanding is that proving this proposition will imply that satisfaction of a wf depends solely on free variables. However I think even if $s$ and $s^′$ share same elements in the same positions, assigning the rest of the different elements of sequences to corresponding bound variables in $\textbf{B}$ may lead to different results. Example:

Let $\textbf{B}$ be $(\forall x_1)(E(x_1) \wedge O(f(x_1))$, where

• $E(x_1)$ is interpreted as "is even"

• $O(x_1)$ as "is odd"

• $f(x_1)$ is "a successor of $x_1$"

• and the domain is a set of positive integers

Here I picked a wf without free variables which should make the initial to-be-proven proposition vacuously true. Now, let $s = (2, \dots)$ and $s^′=(4, \dots)$. We can see that both $s$ and $s^′$ satisfy $\textbf{B}$ (2 and 4 are both even and their successor is an odd integer) just because we were lucky enough to pick the "right" sequences. However if $s^′ = (3, \dots)$, then the wf is not satisfied anymore by $s^′$. As such the proposition saying that "$s$ satisfies $\textbf{B}$ if and only if $s^′$ satisfies $\textbf{B}$" can't be true since $s^′$ fails to satisfy $\textbf{B}$ while $s$ does so.

What am I missing?

Answer 1

$\bf B$ has no free variables hence is satisfied either by every sequence or by none. In fact here: by none, since there are positive integers $n$ for which $E(n)\land O(f(n))$ is false (and even $E(n)$ and $O(f(n))$ are false).

Your $(3,\dots)$ does not satisfy $\bf B$, but neither do your $(2,\dots)$ and $(4,\dots).$

Answer 2

Your formula B is a sentence, it has no free variables. The thereom does hold for it and it says that for any two completely arbitrary sequences $s$, $s'$(that need not match at any place), if $s$ satisfies B then so does $s'$. The thing is that since B is a sentence its truth does not depend on any of the sequences, either it would be true for all sequences or false for all sequences, this follows from induction on wf's and the very definition of satisfaction.