# show that every propositional formula is equivalent to one using the connectives $\to$ and $\neg$

1,363

## Solution 1

As every propositional formula is equivalent to one using only $\land$ and $\neg$, it suffices to prove that $\land$ and $\neg$ can by written using $\to$ and $\neg$ resp. $\mid$.

We have \begin{align*} p \land q &\iff \neg(\neg p \lor \neg q)\\ &\iff \neg(p \to \neg q) \end{align*} and \begin{align*} \neg p &\iff \neg (p \land p)\\ &\iff p \mid p\\ p \land q &\iff \neg \neg (p \land q)\\ &\iff \neg (p \mid q)\\ &\iff (p \mid q)\Bigm| (p \mid q) \end{align*} So we are done.

## Solution 2

The heart of each problem is showing that all of the usual connectives can be obtained using the available ones. I’ll do a bit of each and leave the rest to you.

(a) We know (and you can easily check) that $p\to q$ has the same truth table as $\lnot p\lor q$. Thus, $\lnot p\to q$ must have the same truth table as $\lnot(\lnot p)\lor q$, which has the same truth table as $p\lor q$. That is, using only $\to$ and $\lnot$ I can simulate $p\lor q$ by $\lnot p\to q$. (Of course you should check this by writing out the truth tables!) To get $p\land q$, use de Morgan’s law to get it by combining $\lnot$ and $\lor$, now that you know how to get $\lor$.

(b) By definition $p\mid q$ has the same truth table as $\lnot(p\land q)$. Thus, $p\mid p$ must have the same truth table as $\lnot(p\land p)$, which in turn has the same truth table as $\lnot p$. (Check all this, of course, with the actual truth tables.) Thus, $p\mid p$ simulates $\lnot p$ using only the Sheffer stroke. So what does $(p\mid q)\mid(p\mid q)$ give you? If you don’t see it right away, write out the truth table. Once you have that and $\lnot$, getting the remaining connectives shouldn’t be too hard.

## Solution 3

A start: a) Use a truth table to show that $A\lor B$ is equivalent to $(\lnot A)\rightarrow B$.

b) Use a truth table to show that $\lnot A$ is equivalent to $A \mid A$, and observe that $A\lor B$ is equivalent $(A\mid A)\mid(B\mid B)$.

Share:
1,363

Author by

### Khanak

Updated on June 12, 2022

(a) Show that every propositional formula is equivalent to one using the connectives $\to$ and $\neg$.
(b) Show the same is true for the Sheffer stroke $|$ where $p\mid q$ means "either not $p$ or not $q$" or "not both $p$ and $q$".
You can also use \neg to obtain $\neg$ and \to to obtain $\to$. Choose your favourite.