Verify that $\bigl(p\to(q\to r)\bigr)\to \bigl((p\to q)\to (p\to r)\bigr)$ is a tautology. 
Verify that $\bigl(p\to(q\to r)\bigr)\to \bigl((p\to q)\to (p\to r)\bigr)$ is a tautology.

I am confused on this whole tautology even after looking at examples both in my book and on-line. I started a truth table and this is what I have so far. Can someone please explain this to me in a way that I can understand.
\begin{array}{ccc}
p & q & r & ? & ? & ? & ?\\
\hline
T & T & T\\
T & T & F\\
T & F & T\\
T & F & F\\
F & T & T\\
F & T & F\\
F & F & T\\
F & F & F
\end{array}
 A: A completely different approach (called a tableau proof, and a bit on the informal side for one). Feel free to ignore.

1) Assume $p \implies (q \implies r)$.

2) Assume $p\implies q$.

3) Assume $p$.

4) By (2) and (3) and modus ponens, $q$.
5) By (1) and (3) and modus ponens, $q \implies r$.
6) By (4) and (5) and modus ponens, $r$.

7) By (3)–(6) and implication, $p \implies r$.

8) By (2)–(7) and implication, $(p\implies q)\implies (p\implies r)$.

9) By (1)–(8) and implication, $\bigl(p\implies (q\implies r)\bigr)\implies \bigl((p\implies q)\implies (p\implies r)\bigr)$.

A: The following solution does not use a full truth table, so may not be the intended solution.
How could $\bigl(p\to(q\to r)\bigr)\to \bigl((p\to q)\to (p\to r)\bigr)$ be false?
The sentence  $\bigl(p\to(q\to r)\bigr)$ would have to be true, and the sentence 
$\bigl((p\to q)\to (p\to r)\bigr)$ would have to be false.
How could  $\bigl((p\to q)\to (p\to r)\bigr)$ be false? The sentence $p\to q$ would have to be true, and $p\to r$ would have to be false. To make $p\to r$ false, we must have $p$ true and $r$ false. But if $p$ is true, to make $p\to q$ true, we need $q$ true.
To sum up, the only way that $\bigl((p\to q)\to (p\to r)\bigr)$ can be false if $p$ and $q$ are true and $r$ is false.
However, under these conditions, $\bigl(p\to(q\to r)\bigr)$ is false. 
Thus there is no assignment of truth values to $p$, $q$, and $r$ such that  $\bigl((p\to q)\to (p\to r)\bigr)$ is false and  $\bigl(p\to(q\to r)\bigr)$  is true. 
A: A great way to see results like this is with the help of the method of analytic tableaux. You take the negation of your formula and apply a series of contradiction-hunting rules to get
,
which is closed (meaning that each of its paths end in contradictions), which in turn implies that your original formula is indeed a tautology.
Can you see how this relates to the tables in the other answers?
A: A tautology is a statement that is always true. For propositions $a$ and $b$, $a\rightarrow b$ is logically equivalent to $\neg a\vee b$. If $a=b$, then $\neg a\vee b$ is logically equivalent to $\neg b\vee b$, which is a tautology. So one way to prove that the given statement is a tautology is to consider
$$
a\equiv p\rightarrow\left(q\rightarrow r\right)
$$
and
$$
b\equiv\left(p\rightarrow q\right)\rightarrow\left(p\rightarrow r\right)
$$
and show that $a$ and $b$ are logically equivalent. We can do this by checking the truth table:
\begin{array}{ccccc}
p & q & r & p\rightarrow\left(q\rightarrow r\right) & \left(\left(p\rightarrow q\right)\rightarrow\left(p\rightarrow r\right)\right)\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 1 & 1 & 1\\
0 & 1 & 0 & 1 & 1\\
0 & 1 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 1\\
1 & 0 & 1 & 1 & 1\\
1 & 1 & 0 & 0 & 0\\
1 & 1 & 1 & 1 & 1
\end{array}
(0 is false and 1 is true in the above)
A: Your well-formed formula is an instance of axioms schema 3 of the Łukasiewicz axiomatization of propositional logic.   By the completeness of propositional logic, your wf is a tautology.
