For example, you can replace implication \(\longrightarrow\) with an \(\lor\) and negation:
\[ (A \longrightarrow B) \leftrightarrow (A' \lor B) \]
So we really don't need \(\longrightarrow\), do we? Since we can also write \(\land\) using \(\lor\) and negation,
\[ (A \land B) \leftrightarrow (A' \lor B')' \]
and since equivalence is actually just a pair of implications with an \(\land\), we can conclude that \(\lor\) and negation suffice to write any wff.
Therefore we can conclude that we wouldn't need more than two logical connectives.
But would you believe that a single logical connective will do? Consider the truth table for the Sheffer stroke, |: \[ \begin{array}{c|c|c} {A}&{B}&{A|B}\cr \hline {T}&{T}&{F}\cr {T}&{F}&{T}\cr {F}&{T}&{T}\cr {F}&{F}&{T} \end{array} \] Perhaps you can see that it's a "negated and" (check the truth table): \[ A|B \leftrightarrow (A \land B)' \]
To show that the Sheffer stroke suffices to write any wff, we first show that it can handle negation: \[ \begin{array}{c|c|c} {A}&{A'}&{A|A}\cr \hline {T}&{F}&{F}\cr {F}&{T}&{T} \end{array} \]
Hence
\[ A' \leftrightarrow A|A \]
Furthermore, since the Sheffer stroke is a "negated and", \[ (A \land B)' \leftrightarrow A|B \] we know that \[ A \land B \leftrightarrow (A|B)' \leftrightarrow (A|B)|(A|B) \]
Therefore, since the Sheffer stroke can handle negation, it can reproduce \(\land\); and since \(\land\) and negation suffice to write all of the others (per one of your exercises), the Sheffer stroke alone can write them all!
Isn't that cool? But it would make our language really ugly (maybe we'd get used to it?):
Let
| The apples are not ripe | \(\leftrightarrow\) | The apples are ripe stroke apples are ripe. |
| The apples are ripe and the pie is yummy | \(\leftrightarrow\) | (The apples are ripe stroke the pie is yummy) stroke (The apples are ripe stroke the pie is yummy). |
Etc.! We'd definitely need a verbal cue for parentheses....:)
N.B.: I'm fond of saying that you can't negate an implication with an implication, because we need to flip all the T to F, and the F to T; and so we need to negate the "truthfully promiscuous" implication with something "falsefully promiscuous" (i.e. and). But clearly, this is only informally true, since the "truthfully promiscuous" Sheffer stroke must be negated by a Sheffer stroke!