The question I have on my homework is
"Find a formula that contains no connective other than ¬ and v which is equivalent to ((p→q)→s)."
I drew out the truth table for the given formula but have no idea where to even start to construct an equivalent formula.
$$\begin{array}{c} p&q&r&\text{((p→q)→s)}\\ \hline 1&1&1&1\\ 1&1&0&0\\ 1&0&1&1\\ 1&0&0&1\\ 0&1&1&1\\ 0&1&0&0\\ 0&0&1&1\\ 0&0&0&0 \end{array}$$
Thanks for any help
The game is to replace $\rightarrow$ by $\neg$ and $\lor$ using the conditional law. Along with DeMorgan's law to swap $\lor$ for $\land$ (and vice versa), you can get expressions using just $\lor$, just $\land$, or just $\rightarrow$, as desired (with $\neg$ always possibly involved).
In case you are not aware of the conditional law, it is that $p \rightarrow q$ is equivalent to $\neg p \lor q$.
We use, twice, the identity $$a\rightarrow b \equiv \lnot a \lor b$$
$$\begin{align} (p\rightarrow q) \rightarrow s &\equiv \lnot(p \rightarrow q) \lor s \\ \\ & \equiv \lnot(\lnot p \lor q) \lor s\\ \\ \end{align}$$
Remark: Every propositional statement can be expressed in terms of only $\lor$ and $\lnot$.
