Trouble with a theorem in Hunter's Metalogic I'm a logician studian and I'm reading Hunter's Metalogic. I'm having trouble understanding and exemplifing part of a theorem in the book. It's the theorem 40.14, pp. 156-7. 40.14. Let t and u be terms. Let t' be the result of replacing each occurrence of vk in t by u. Let s be a sequence, and let u*s=d, i.e., let the member of D assigned by I to u for the sequence s be d. Let s' be s(d/k): i.e., let s' be the sequence that results from substituing d for the kth term of s. Then t' _s=t_ s': i.e. the member of D assigned by I to t' for the sequence s is the same as the member of D assigned by I to t for the sequence s'. Proof (by induction): Basis: n=1 There's 3 cases: 1. $t$ is a constant; 2. $t$ is a variable $v_j$ and $j$ is different from $k$ 3. $t$ is a variable $v_j$ and $j=k$. Cases 1 and 3 are easy. But I cannot prove case 2, and also I cannot think of an example of it. Can anyone help? Thanks!

The _assignment_ function $s$ is a function :

> $s : Var \to D$

where $Var$ is the set of _variables_ : $v_1, v_2, v_3, \ldots, v_k, \ldots$, and $D$ is the _domain_ of the _interpretation_ $I$.

Assume a simple example with $D = \\{ a, b, c, d, ... \\}$ and let $s$ the following assignment :

> $s(v_1)=a, s(v_2)=b, s(v_3)=c$ and $s(u)=d$.

Consider $s'=s[d/2]$, i.e. :

> $s'(v_1)=a, s'(v_2)=d, s(v_3)=c$.

In this case, we have : $k=2$.

What happens if $t$ is $v_3$, i.e. $t$ is $v_j$ with $j=3$ and thus $j \
e k$ ? Clearly, $v_2$ does not occur in $t$; thus, the replacement of each occurrence of $v_2$ in $t$ by $u$ does not produce any change and $t'=t$.

As you can see : $s'(v_3)=c=s(v_3)$ and thus $s'[t]=s'(v_3)=c=s(v_3)=s[t]=s[t']$.