# Curious tautological pattern on"p->p"

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:

1. "((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

• Check this by subsituition $$\color{red}{p\to p}\equiv\top\tag{1}$$ $$\boxed{\color{red}{(p\to p)}\to p}\equiv\top \to p\equiv p\tag{2}$$ $$\underline{\boxed{(\color{red}{(p\to p)}\to p)}\to p}\equiv p\to p\equiv \top\tag{3}$$ $$(\underline{\boxed{(\color{red}{(p\to p)}\to p)}\to p)}\to p\equiv\top\to p\tag{4}$$ $$\vdots$$ May 29, 2020 at 20:34
• An astute observation. You have a good eye :) Jun 3, 2020 at 4:00

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.

$$(p \rightarrow p) \rightarrow p$$ 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 $$p$$.
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 $$p \rightarrow p$$.