How are theorems tautologies? I am reading an analysis book that introduces techniques of proofs, and there is a part that I am not understanding. I would appreciate some elaboration on what is being said. The paragraph in question goes as follows:
"The most common type of mathematical theorem can be symbolized as $p \implies q$, where $p$ and $q$ may be compounded statements. To assert that $p \implies q$ is a theorem is to claim that  $p \implies q$ is a tautology; that is, that it is always true. We know that $p \implies q$ is true unless $p$ is true and $q$ is false. Thus, to prove that $p$ implies $q$, we have to show that whenever $p$ is true it follows that $q$ must be true."
Okay, everything in this paragraph is fine with me except for the second sentence: "To assert that $p \implies q$ is theorem is to claim that  $p \implies q$ is a tautology; that is, that it is always true." How can this be true if a tautology is something that is always true for all cases, and just as it was mentioned in the paragraph, $p$ implies $q$ is true in all but one case-- that is when $p$ is true and $q$ is false. This sentence seems contradictory since it contradicts the definition of a tautology, could anyone elaborate? Also, I don't know a lot about logic, just basic propositional and predicate logic-- thank you.
 A: Note that $p$ and $q$ in this quote are not propositional variables, but placeholders for more complex formulas. This means that, depending on which particular formulas they are, the truth value of each of $p$ and $q$ can depend on the value of various variables, and it is possible for one or more of these variables to appear in both $p$ and $q$. When that is the case, it may well be that every assignment of values to the actual variables gives a combination of truth values for $p$ and $q$ that makes $p\Rightarrow q$ evaluate to true.
For example, suppose that $p$ and $q$ are both the (very simple) formula that is a single propositional variable $A$. Then what $p\Rightarrow q$ means is the actual formula
$$ A \Rightarrow A $$
which is a theorem because every choice of value for the variable $A$ leads $A\Rightarrow A$ to be true.
Or suppose $p$ is the formula $A\land B$ and $q$ is the formula $B\lor C$ (where $A$, $B$, $C$ are propositional variables). Then $p\Rightarrow q$ means the formula
$$ A\land B \implies B\lor C$$
which is again a tautology, because every combination of values for $A, B, C$ makes this come out true.
On the other, hand if $p$ is $A$ and $q$ is $B$, then we're looking at
$$ A \implies B $$
which is not a tautology (and not a theorem), because it can be made false by the truth assignment where $A$ is true and $B$ is false.
A: The theorem only holds if the case in which it could be false never happens. Let's look at a concrete example. Let $p$ represent the statement "The legs of a right triangle have lengths $a$ and $b$, and the hypotenuse has length $c$", and let $q$ represent the statement "$a^2+b^2=c^2$".
Now, what cases exist here? We can have $p$ and $q$ both true, we can have $p$ and $q$ both false, and we can even have $p$ false while $q$ is true (for example, $a=b=c=0$). However, there does not exist an actual case where $p$ is true but $q$ is false. The non-existence of a case that would ruin the tautology is precisely what makes the theorem true. Showing that such a case cannot exist amounts to proving the theorem.
Does this answer your question?
