Proof by induction of inadequacy of set of propositional connectives So we have to show that given a particular set of propositional connectives that the set is not adequate. I am comfortable with what a set being adequate means but I can't get my head around why the proof by induction works and is sufficient. This is a question and solution which is given in my notes:
"Q) Prove that the connective § with truth table shown below is not adequate.
$$\begin{array}{cc|c}
p & q & p§q \\ \hline
T&  T&  F\\
T&  F&  F\\
F&  T&  T\\
F& F & F\\
\end{array}$$
A) We prove by induction on complexity of terms, that for any term built
up from a set L of propositional variables using § only - let us denote the set of
such terms by Term§(L) - and for any valuation v, if v(p) = F for every p ∈ L,
then v(s) = F for every s ∈ Term§(L). That will be enough since then there
will be no term in Term§(L) which is a tautology.
Base case (s a propositional variable): this is our hypothesis.
Induction step (just one): Suppose that s = t§u where, by induction, we
may assume that v(t) = F = v(u). Then, consulting the truth table for §, we
see that v(s) = F, as required."
I understand why the result shows that the set is not adequate as it shows that there is no term that can be built from the connectives that make a tautology, however I don't understand the assumption that: 
"If v(p) = F for every p ∈ L, then v(s) = F for every s ∈ Term§(L)"
How can we just make this assumption? What about the valuations where v(p) = T? Don't they need to be accounted for? Struggling with the whole idea with the proof on complexity on terms so thanks for reading.
 A: *

*I think you agree that  in order to show that  § is inadequate, it is enough to show that there is no way to express a tautology with just  §.

*A tautology is a term $t$ whose value $v(t)$ is true for every assignment of values to the variables of $t$.

*If there is any assignment of values to $t$'s variables that results in $t$ having a false value, then $t$ is not a tautology. 

*In particular, if assigning false values to  all the variables of $t$  results in $t$ having a false value also, then $t$ is not a tautology. (This is the crucial point that seems to be giving you trouble.) 

*But it is the case that if $t$ is made of only §, then assigning false values to all its variables results in $t$ also having a false value. (This is the step that is justified by the induction proof.)

*Therefore, any term $t$ made of only § is not a tautology.

*Therefore, § is not adequate.
Is this clearer now?

You said you don't understand how to go from step 3 to step 4.
Consider this silly example, which follows the same pattern:


*

*A paradise is a country where every inhabitant is happy.

*Gaston, who lives in France, is not happy.

*Therefore, France is not a paradise.


“But,” you say, “why don't we have to consider Pierre?  Pierre also lives in France!”
Pierre doesn't matter.  France is only a paradise if every person in France is happy.  To show that France is not a paradise, we only have to find one person in France who is not happy.


*

*A tautology is a term for which  every assignment of values to its variables results  in its having a true value.

*Assigning all false values to the variables of the term $t$ results in $t$ having a false value.

*Therefore, $t$ is not a tautology.


“But,” you say, “why don't we have to consider other assignments of values to the variables of $t$?  Those are assignments too!”
Those assignments  don't matter.  $t$  is only a tautology if every assignment results in $t$ having a true value.  To show that $t$ is not a tautology, we only have to find one assignment that results in $t$ having a false value.
A: You must first show that the Basis step holds.
Whay does it mean ? That a formula built up with the § connective with only one propositional variable $p$ can never be a tautology.
This is because $p§p$ must be only $True-True$ or $False-False$, and both are mapped into $False$ by the truth-table.
Having proved this, you must use it in the Induction step that "reduce" the case $\alpha § \beta$ to the verification (via the truth-table) that the property hold for it, assuming that it holds for $\alpha$ and $\beta$ (being "shorter", the induction hypotheses applies to them). 
A: Your set of terms T, I take it, is the smallest subset of the set of strings, S, generated from your propositional variables P, and the symbols §, (, and ) using the production 
s,t in T => (s§t) is in T.  (1)
That is, T is the smallest subset of S containing P and closed under (*).
Now if we have a valuation of v: P --> {T,F} which maps every element of P to F, then let 
TF be the subset of T on which v has value F, that is,
let TF = {t in T | v(t) = F}.  Now by our choice of v, P is contained in TF.
Further if s,t are in TF then by the truth table definition of §, (s§t) is in TF. Hence TF contains P and is closed under (1). But your set of terms that I have called T is the smallest subset of S containing P and closed under (1).  Hence TF = T and indeed T contains no tautologies.   
