This question is from Herbert B.Enderton, A Mathematical introduction to Logic.
Section 2.2; question 13
Homomorphism Theorem: Let $h$ be a homomorphism of $U$ into $B$, and let
s :$V \rightarrow |U|$ where $V$ is the set of variables.
Prove that, for any term $t$, we have $h(\bar{s}(t)) = \overline{h\circ s}(t)$, where $\bar{s}(t)$ is computed in $U$ and $\overline{h\circ s}(t)$ is computed in $B$.
My solution:
Case 1: $t$ is a constant symbol
We have $h(\bar{s}(t)) = h(c^U) = c^B$ since $h$ is a homomorphism
$c^B = \overline{h\circ s}(t)$. Done.
Case 2: $t = f_{t_1........t_n}$ where $f$ is a $n^{th}$ place function symbol
$h(\bar{s}(t)) = h(f^{U}(\bar{s}(t_1),...,\bar{s}(t_n))) = f^B(h\circ\bar{s}(t_1),.........,h\circ\bar{s}(t_n))$, since $h$ is a homomorphism.
$f^B(h\circ\bar{s}(t_1),.........,h\circ\bar{s}(t_n)) = \overline{h\circ s}(t)$
Q.E.D
My question is now the following.
The books states that we should prove the theorem by induction on $t$, but in my attempt I did'nt really use induction (as you can see from the absences of an Induction hypothesis). So is my proof wrong, if so please point out my mistake and perhaps guide me in how to use induction to complete the proof.
P.S: My apology for those who have the book, I do not know how to typeset the weird looking U and B in the book