On Peirce’s law and the law of the excluded middle

A few days ago I found a video on YouTube explaining how to use CoqIde, the IDE for the Coq proof assistant. The proof that was used to illustrate the IDE was the equivalence between Peirce’s law and the excluded middle law, that is, the equivalence between the two following laws:

$\mleq{(0)}
{((p{\Rightarrow}q){\Rightarrow}p)\Rightarrow p}
$

and
$\mleq{(1)}
{\neg{p} \vee p}
$

After watching the video, I wrote a goal-oriented, calculational proof of the equivalence between (0) and (1). In this short note, I describe the design of my proof.

A calculational proof

A useful rule of thumb is to start with the most complicated side, for it is usually easier to simplify expressions than to complicate them. With this in mind, let us try to manipulate (0) so that it becomes (1).

Looking at the shapes of (0) and (1), we immediately see that $q$ occurs in (0), but not in (1). So, if we start with (0), we will have to eliminate the occurence of $q$. This suggests that we have to use laws that involve the elimination of variables. Two elementary rules that are of this kind are the so-called absorption rules for conjunction and disjunction:

$\mleq{(2)}{(p \wedge q) \vee p ~\equiv~ p}$

$\mleq{(3)}{(p \vee q) \wedge p ~\equiv~ p}$

At a first sight, these rules may not seem very useful, because they involve only conjunction and disjunction and we want to eliminate a variable from a formula involving implication. But we know that implication can be written in terms of negation and disjunction:

$\mleq{(4)}
{{p{\Rightarrow}q} ~\equiv~ {\neg{p} \vee q}}
$

These observations, together with the De Morgan rule, allow us to write a goal-oriented, calculational proof of the equivalence between (0) and (1):

\[
\beginproof
\pexp{((p{\Rightarrow}q){\Rightarrow}p)\Rightarrow p}
\hint{=}{rewrite the two leftmost $\Rightarrow$ using (4)}
\pexp{{\neg{(\neg{p} \vee q)} \vee p} ~\Rightarrow~ p}
\hint{=}{De Morgan rule, that is, $\neg{(\neg{p} \vee q)} ~\equiv~ (p \wedge \neg{q})$}
\pexp{{(p \wedge \neg{q}) \vee p} ~\Rightarrow~ p}
\hint{=}{absorption (2)}
\pexp{p{\Rightarrow}p}
\hint{=}{rewrite $\Rightarrow$ using (4)}
\pexp{\neg{p} \vee p ~~.}
\endproof
\]

The video

The video that I mentioned is shown below. It would be unfair to compare the proofs, since the main goal of the video is to illustrate Coq’s IDE and not to show how to design proofs. Nevertheless, I would like to invite the reader to comment on the use of automated/interactive theorem provers to design proofs. Thanks for reading.

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