Confused by Rosen's explanation of Proof by Contradiction

Question

Suppose we want to prove that a statement $p$ is true. Furthermore, suppose that we can find a contradiction $q$ such that $\neg p \rightarrow q$ is true. Because $q$ is false, but $\neg p \rightarrow q$ is true, we can conclude that $\neg p$ is false, which means that $p$ is true.

Chapter 1, page 86 of K.H. Rosen's Discrete Mathematics and Its Applications (7th ed)

How can we show $ \neg p \rightarrow q$ is true? We'd have to show that if $\neg p$ is true, then $q$ is true (because this rules out the only case where the implication is false) - but the latter is impossible, as a contradiction is always false.

Maybe we don't need to show that the above implication is true?! Instead, can we argue that the above implication is true due to the fact that we've only used valid arguments to arrive at $q$? If so, how exactly can we formalizes this argument?

To summarize my question: How does Rosen conclude that the implication $\neg p \rightarrow q$ is true?

Answer 1

How can we show $¬p→q$ is true?

Usually by a conditional subproof. Whenever under the assumption of $\neg p$ you may derive $q$, then you may deduce $\neg p\to q$.$$\begin{array}{|l}\quad\begin{array}{|l}\neg p\\\hline \vdots\\ q\end{array}\\\neg p\to q\end{array}$$

How you derive $q$ in that subproof will depend on what else you have to work with.

... but the latter is impossible, as a contradiction is always false.

A contradiction can be derived from an inconsistent set of assumptions, because they are inherently contradictory. Programmers have an acronym for this: G.I.G.O., or Garbage In Garbage Out.

Deriving something under an assumption means that it will be true on condition that the assumption is true also (well, that and all prior assumptions). So if the conclusion is a contradiction, then the assumptions are suspect.

That's the basis of Proof by Contradiction: If the assumption produces garbage, then throw it out.$$\begin{array}{|l} \quad \lnot p\to q\\\quad \lnot q\\\therefore~p\end{array}$$

Answer 2

However, how can one show that ¬p→q is true? We'd have to show that if ¬p is true, then q is true (because this rules out the only case where the implication could be false) - but the latter is impossible, as a contradiction is always false.

This isn't how the system works. By this criterion, one would have to reject the idea that $q → q$ always holds, because if $q$ is actually false, the same sort of argument applies.

$p → q$ being true doesn't imply that $q$ is true. It means that $q$ is at least as true as $p$. When $p$ is false, $q$ may be false as well, because false is just as true as false. The flip side of this is that $p$ can be at most as true as $q$. So if $q$ is false, then $p$ must be false, because false is minimally true.

Edit: Perhaps this will help. It seems like you might be getting stuck on some kind of informal semantic reasoning that proving an implication requires reasoning from "$p$ is true" to "$q$ is true." But $q$ is false, and therefore "cannot be true." However, at this semantic level, $p$ could be false, so that the premise is "false is true." Given that "false is true" it's no problem to conclude "false is true." So it is erroneous to presume that because we know "$q$ is false," there are no premises that yield "$q$ is true."

Also, it seems like what you're reading has not adequately conveyed the relationship between proof and truth.

• Proving is an activity where you use inference rules and axioms to deduce propositions from other propositions. A theorem is a proposition that can be deduced entirely from axioms (or similar).

• Truth involves assigning values to propositions by choosing values for each atomic proposition, while the values of compound propositions are functions of the values of their sub-formulas.

There are meta-theorems relating proof and truth. For example, "a proposition is a theorem if and only if it is true under every assignment of atomic propositions." And the inference rules are designed to 'preserve truth.'

But when you 'prove' something, it means you are using the rule system. The meta-theorems mean that based on thinking about truth, one can conclude that a proof could be built. But inspecting truth values and such is not the proof.

Answer 3

I think I've understood it now: We start from $\neg p$ and some other assumptions $\varphi_1,\ldots,\varphi_n$ which we know are true. We then make valid arguments (e.g. by using inference rules) which lead to a contradiction $q$. And the argument is valid if and only if $(\neg p \wedge \varphi_1 \wedge \ldots \wedge \varphi_n) \rightarrow q$ is a tautology. The only way for this implication being true is that $\neg p$ is false, i.e. $p$ is true.

So the implication in the quote is true because we used a valid argument.

Answer 4

I will admit it's a little difficult to understand what your question based on reading your question and the comments, but I'll at least throw in my two cents here to see if I could help out (if you get completely lost by it, feel free to say so).

How proof by contradiction works is that a proposition is either true of false. We get a contradiction when we show a statement is both true and false, showing that our initial assumptions are inconsistent. For implication statements, we use this to show $P \implies Q$ by assuming both $P$ and $\neg Q$ are simultaneously true and deriving a contradiction. When we get this contradiction, it means one of our assumptions cannot be correct.

In general, let's say we want to prove a proposition $P$. To prove it by contradiction, suppose $\neg P$. Then we write some sentences and eventually we'll get $R$ and $\neg R$ where $R$ is another statement. Since that last line of reasoning is inconsistent, this ends the proof because after all, $\neg P \implies (R \text{ and } \neg R)$ is true (and you can use a truth table to verify that).

Notice how if you have a false statement and another false statement, then a false statement implying another false statement makes the entire implication true. For instance, $\neg P$ is a false statement and $R \text{ and } \neg R$ is a false statement. By a truth table, $\neg P \implies (R \text{ and } \neg R)$ is true. That's basically the idea behind proof by contradiction.

Why exactly is a false statement implying another false statement considered "true"? Let's use an example: "If I have 60 dollars ($P$), then I am able to buy a new video game for 60 dollars ($Q$)." If I don't have 60 dollars ($\neg P$), then obviously I can't buy a new video game for 60 dollars ($\neg Q$) because I don't have enough money. Both of those implication statements are true, and we can extend that same idea into mathematics as well.

Answer 5

How can we show $ \neg p \rightarrow q$ is true? We'd have to show that if $\neg p$ is true, then $q$ is true (because this rules out the only case where the implication is false) - but the latter is impossible, as a contradiction is always false.

$q$ can tentatively be true and false at the same time, but if you want to continue with your axioms being consistent, you need to reject the assumption that caused this contradiction (of $q$ being true and false at the same time). In this case, the assumption that caused/led to the contradiction was "$\neg p$ is true".

If you never allowed for the tentative (conditional) possibility of a statement being true and false at the same time, then there would be no such concept as a contradiction...

Answer 6

Suppose we want to prove that a statement $p$ is true. Furthermore, suppose that we can find a contradiction $q$ such that $\neg p \rightarrow q$ is true. Because $q$ is false, but $\neg p \rightarrow q$ is true, we can conclude that $\neg p$ is false, which means that $p$ is true.

Let me rephrase what the author wrote, hopefully clearer:

• Suppose that we can determine that $(\neg p \rightarrow q)$ is valid and that $q$ is a contradiction. This means that $q$ is false, which means that $\neg p$ is false, which means that $p$ is true.

  In other words, if by assuming the negation of statement $p$ then performing a sequence of valid steps/arguments we can derive a contradiction, then we know that $p$ is true.

To summarize my question: How does Rosen conclude that the implication in the quote is true?

Maybe we don't need to show that the above implication is true?! Instead, can we argue that the above implication is true due to the fact that we've only used valid arguments to arrive at $q$?

Precisely! $$\neg p \rightarrow q$$ is an argument with premise $\neg p$ and conclusion $q.$ It is valid (not merely true) precisely because in making it, you have performed only valid steps/ sub-arguments.