Proof by contradiction, why does it work?

Question

I'm reading learning to reason by Nancy Rodgers, and in her textbook she says: (we have to derive $r$)

"To justify the validity of a proof by contradiction, we can argue as follows. When we assume $∼r$ is true and derive c and -c, we have proved the following implication: $$∼r⇒(c∧∼c)$$ Next we translate the above implication as an or-sentence. $$r∨(c∧∼c)$$ since $(c∧∼c)$ is false, $r$ must be true."

I dont really get this, "assume $∼r$ is true" blah blah.. "so $r$ must be true" how can $∼r$ and $r$ both be true?

One more doubt, I know she used this equivalence: $(p∨q)⇔ (∼p⇒q)$, but I dont get this either:
"it's blue or it's red $⇔$ if it's not blue then it's red" what if it's yellow though? both sentences would be false (correct since they're equivalent) but I can't say that since "it's not blue" it's red, because it's yellow! This is what the the textbooks seem to say: "since $(c∧∼c)$ is false, $r$ must be true." but how? what if it's another thing? I'm so confused, can you guys help me please?

Answer 1

Without writing symbols, the concept is simple: we want to show that $r$ is true. If we assume $r$ is false, then something impossible happens, which means our initial assumption was wrong, so $r$ must be true. Note that "not false" is the same as "true."

With symbols, we begin by assuming "not $r$", which is written $\neg r$. By "something impossible," we mean a contradiction, which is when a statement is both true and false. A contradiction is thus written $c\wedge \neg c$ for some $c$. Then the substance of our proof is the implication $$\neg r\implies (c\wedge\neg c).$$

We can interpret any implication $a\implies b$ as the truism "either $a$ is true, in which case $b$ is true, or $a$ must be false." Thus, no matter what, either $\neg a$ is true or $b$ is true, so we can write $\neg a\vee b$.

Going back to $\neg r\implies (c\wedge \neg c)$, we then have the true statement $\neg(\neg r)\vee (c\wedge\neg c)$, which is the same as $r\vee(c\wedge\neg c)$. Thus either $r$ is true or $c\wedge \neg c$ is true. The latter is clearly false, so we must have that $r$ is true.

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Regarding the second question: the fact we used is that $$(a\implies b)\implies (\neg a\vee b),$$

and indeed it is true that

$$(a\implies b)\iff (\neg a\vee b).$$

To see the other direction, suppose $\neg a\vee b$. If $a$ is true, then $\neg a$ is false, so $b$ must be true, whence $a\implies b$.

In your example, you have $(p\vee q)\iff (\neg p\implies q)$, where $p$ is the statement "it is red" and $q$ is the statement "it is blue" ("it" being some object). Suppose it is red or blue. Then if it is not red, it is blue. (This corresponds to $(p\vee q)\implies (\neg p\implies q)$.) Now suppose that, being not red implies it is blue. Either the object is red (in which case $p\vee q$ is true), or it is not red, in which case it is blue, so $p\vee q$ is still true.

Note that if the object is yellow, then $p\vee q$ is not true, so the statement $(p\vee q)\implies (\neg p\implies q)$ is vacuously true.

Answer 2

One thing just to be clear: $\sim A$ means 'not $A$' right? I am more familiar with the notation $\neg A$.

Basically you already answered your own quastion:

She assumes, that $\neg r$ is true and then she finds out (this is the core of a proof by contradiction) that "$\neg r \Rightarrow (c \wedge \neg c)$ is also true, which again is - as you stated - EQUIVALENT to $r \vee (c \wedge \neg c)$. Since this last expression must then be true, at least either $r$ or $(c \wedge \neg c)$ must be true. But $(c \wedge \neg c)$ is never true - nothing can be red and not red at the same time.

Your counterexample in the end is no counterexamble, since the equivalence still holds as you wrote yourself.

"it's blue or it's red $\Leftrightarrow$ if it's not blue then it's red" what if it's yellow though?"
But if you know "it's blue or it's red" to be true, then the implication "if it's not blue then it's red" is correct.

Answer 3

In the first case when you assume $\lnot r$ is true, you don't know it is true it may or may not in actuality be true.. But in presuming it is true you do prove that $\lnot r \implies \lnot c$ is true, always true.

Consider the possibility $n^2 = 4k + 2$ for some integer $k$. We don't know if it is true or not. And we are not claiming it is true but we are considering what may happen if it is true.

But if it is true $n^2 = 2(2k+1)$ is even so $n$ is even. So $\frac n2$ is an integer and $n*\frac n2$ is an even integer. And $n*\frac n2=\frac{n^2}2 = 2k+1$ is an even integer.

So we have proven that $n^2 = 4k + 2$ for some integer $k \implies 2k+1$ is even. The implication is a true statement whether or not $n^2$ does or does not equal $4k +2$ for some integer $k$.

And no we know that $2k+1$ is odd is always true.

So you have $n^2 = 4k + 2\implies 2k+1$ is even. And we have $2k+1$ is odd. Then $n^2 = 4k + 2 \implies (2k + 1$ is even $\land 2k+1$ is odd$)$.

That is true whether are not we know that $n^2 = 4k +2 $ or not.

But $(2k+1$ is even $\land 2k+1$ is odd$)$ is certainly false.

So the only way that $n^2 = 4k + 2 \implies (2k + 1$ is even $\land 2k+1$ is odd$)$ is true is if $n^2 = 4k + 2$ is false.

So $n^2 = 4k + 2$ is false. We never asssumed it was true but we showed if it were true a contradiction would arise.

.......

"it's blue or it's red ⇔ if it's not blue then it's red" what if it's yellow though? both sentences would be false (correct since they're equivalent)

Exactly. Nothing wrong with that. "it's blue or it is red" is false and $"if it's not blue then it's red" is false.

Equivalency can hold if both are false.

but I can't say that since "it's not blue" it's red, because it's yellow!

Of course not. "if it's not blue it's red" was a FALSE state.

"If an animal isn't a horse, then it is a pig" is false. So you show me a rabbit. And I claim, since it's not a horse it's a pig.

Why would I do that. "If an animal isn't a horse, then it is a pig" is FALSE so I wouldn't say that.

This is what the the textbooks seem to say: "since (c∧∼c) is false, r must be true." but how? what if it's another thing?

Other than what?

Let's suppose it is yellow. Then we know it is not red and we know it is not blue.

And we proved "if it's either red or blue then if it's not blue it is red" That would mean if (it's red or blue) then as it's not blue (it is red)

So (it's red or blue)$\implies$ (it's red).

And we know (it's not red).

So (it's red or blue) $\implies$ (it's red and it's not red).

As the later is certainly false then (it's red or blue) must be false. And it IS false. It's yellow!

Answer 4

The proof by contradiction is simply based on the fact that :

If the conclusion of a Valid Argument is False ( as a contradiction of the form $\lnot{c}\wedge c $ for example), then we are sure that its Premisse is also False.

Answer 5

I dont really get this, "assume $∼r$ is true" blah blah.. "so $r$ must be true" how can $∼r$ and $r$ both be true?

She is not saying that $\neg r$ and $r$ are both true. Rather, the proof by contradiction shows that if you assume $\neg r$ to be true, you get a contradiction. That is: If $\neg r$, then we get a contradiction. But since we can't allow to have any contradictions, that means that you cannot have $\neg r$ either. Thus, $\neg r$ is in fact false ... and thus $r$ has to be true.

Think of it like this. Suppose we say "If we do X, then all hell will break lose. Therefore, we better not do X"

One more doubt, I know she used this equivalence: $(p∨q)⇔ (∼p⇒q)$, but I dont get this either:
"it's blue or it's red $⇔$ if it's not blue then it's red" what if it's yellow though?

Sure, if it's yellow, then we can't say that if it's not blue, then it must be red. However, the proof established with certainty (remember: it's a proof!) that if $\neg r$, then you get a contradiction. So given it has established that, there really is no third 'yellow' option anymore: either $r$ is true, or the contradiction is true. But, obviously, the contradiction ('blue') is not true, and so $r$ ('red') is the only option left.