# Whis is $((p\implies q)\implies q)\implies q$ not a tautology?

I am studying logic and I am having difficulty understanding how $$((p\implies q)\implies q)\implies q$$ (where $p,q$ are some predicates) is not a tautology.

My reasoning is that if $p$ is true then it follows that the rest is also true(if $q$ is false I would have a contradiction about $p$ being true),while if $p$ is false I have again that it implies the rest since false implies anything.

My definition of a tautology is a proposition that's true indipendently of the fact that $p,q$ are true or false;so following this definition I've shown that the above formal construct holds whether $p$ is true or not.

What am I missing ?

• Try proving that $((p\to q)\to q)\to q$ is equivalent to $p\to q$. – egreg Mar 4 '16 at 15:13
• @egreg Intuitively I could say that the implications after the first $(p \to q)$ do not change the initial condition $p \to q$ since they follow from it ,but formally I am clueless about how to show that.Should I make a truth table for both $((p \to q)\to q)\to q$ and $p\to q$ and show that for each choice of true or false for $p,q$ they're the same (or is there a simpler argument)?Thanks for your advice. – Mr. Y Mar 4 '16 at 15:24
• I'm not saying this is preferable to the accepted answer, but it can shed light on your understanding of propositional calculus. A truth table is not difficult to build. Or with Boolean computations. – egreg Mar 4 '16 at 15:26

Consider $p$ as true and $q$ as false.
Then $p \implies q$ is false.
$false \implies q$ is true.
$true \implies q$ is false.
Remember that a tautology is a statement that outputs true regardless of the inputs. You are required to set values for both $p$ and $q$, not just $p$.