Curious tautological pattern on"p->p"

Question

I found something that boggles me ( I'm really a beginner in symbolic logic, so maybe it's very trivial).

I was practicing with truth-tables, and I found that:

1. "p->p" is a tautology

2. "(p->p)->p" is not a tautology.

I decided to go further, and:

3. "((p->p)->p)->p" is again a tautology, but

4."(((p->p)->p)->p)-> p" is not, and it keeps alternating.

I checked with an online logic calculator, and it seems correct.

Now, do you know why is that? Is there any particular reason for this pattern?

Cheers

Answer 1

1) For any formula $p$, $p \to p$ is a tautology.
2) For any tautology $T$, $T \to p$ is logically equivalent to $p$. (Check it out with a truth table.)

So an even amount of occurrences of $p$ will give you tautologies (by 1)); appending another $p$ will give you something that behaves like p (by 2)). And if you take that $p$-equivalent proposition and append another $p$, you will, by 1), get the tautology again, etc.

The core answer to your question lies in 2).

Answer 2

I love @lemontree's explanation (and @Manx's terser version of the same point). It may be useful to see the point in a different way, though:

$(p \rightarrow p) \rightarrow p$ is equivalent to $\neg (\neg p \lor p) \lor p$, which by one of De Morgan's laws is equivalent to $(\neg \neg p \,\&\, \neg p) \lor p$.

That last expression is equivalent to $(p \,\&\, \neg p) \lor p$.

The left disjunct of that expression is a contradiction, though; it's always false, so the truth value of the whole expression is the same as the truth value of $p$.

lemontree's answer shows that you can apply this conclusion to understand the longer expressions.

For example, $((p \rightarrow p) \rightarrow p) \rightarrow p$ embeds the original 3-$p$ expression on the left side of the arrow, and we saw that that was equivalent to $p$, so the whole expression is equivalent to $p \rightarrow p$.