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To show that an argument is valid, why can we not assume that both its premises and conclusion are true then show that there's no contradiction?

  • Example:

    If $x^2=4$ and $x\neq-2,$ then $x=2.$

    Proof

    Suppose that $x=2$ (assuming the conclusion).

    Then $x^2=4$ and $x\neq-2.$

    Thus, if $x^2=4$ and $x\neq-2$, then $x=2.$

With proof by contradiction, it is legitimate to assume that the conclusion is false; why?

ryang
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enoopreuse22
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    That way you could also prove "Suppose $x^2 \ge 4$ and $x\neq-2$. Then $x=2$". – dxiv May 12 '23 at 19:06
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    Suppose $\tan(x) = 1$ and $x\ne -3\pi/4$. Then $x = \pi/4$. You won't derive a contradiction from these statements, but $x = \pi/4$ could still be false. (It could perhaps be $5\pi/4$.) – eyeballfrog May 12 '23 at 19:11
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    You've proved that if $x = 2$, then $x = 2$. That's probably not what you're going for. – anomaly May 12 '23 at 19:58
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    If this worked, you could prove a stronger statement: If $x^2=4$ the $x=2.$ Proof: Suppose $x=2$. Since $x=2$, we have $x^2=4$. QED. That proof follows the exact same structure as yours and is just as valid (which is to say, not at all valid). – David K May 13 '23 at 04:21

4 Answers4

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Suppose that I promise that if you sweep my porch, then I will give you a reward, and furthermore I promise that the reward will not be an old pair of sneakers.

“Maybe the reward will be a bag of gold!” you say to yourself. “Suppose the reward is a bag of gold! MJD said he would give me a reward for sweeping his porch, but it would not be a pair of old sneakers. A bag of gold is a reward, and it is not an old pair of sneakers. MJD must be going to give me a bag of gold!”

Then you sweep my porch and I give you a pack of gum. “Here's your reward,” I say.

You wanted a bag of gold. “I proved logically that you were going to give me a bag of gold!” you complain.

“Well,” I reply, “it seems that your proof was wrong.”

MJD
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I really laughed when reading MJD's excellent answer.

However you may want a more explanatory answer.

When you prove that "$A$ and $B$ are not in contradiction", what you really prove is that it cannot be proven that $(A \land B)$ is false (without any additional hypothesis).
Note that this does not prove that $(A \land B)$ is true: if $A$ or $B$ include some free variable $x$, and you have proven that it cannot be proven that $(\forall x, A \land B)$ is false, it may just mean that for some values of $x$, $(A \land B)$ is false, but it is true for other values of $x$.
If you have proven that $(\forall x, \text{it cannot be proven that } (A \land B) \text{ is false})$, then, assuming your theory is complete, you could deduce that $(\forall x, (A \land B) \text{ is true})$.

But anyway, in the example you present, you do not at all prove that the premises and conclusion are not contradictory. To prove that, you would need to prove that all possible logical reasonings from those hypotheses never reach a contradiction. There is an infinity of possible reasonings, so the arguments required are much, much more complex than what you wrote.

In proof by contradiction, the situation is different. We want to prove $A \Rightarrow B$. So we assume $\lnot B$, and we prove there is a contradiction with $A$. This is much easier than proving that there is no contradiction. You just have to find one contradiction, not prove that all reasonings never reach a contradiction.
Then, assuming the theory is consistent, this proves that $A \land \lnot B$ is false, so its contrary, $\lnot A \lor B$ is true. And this is logically equivalent to $A \Rightarrow B$.

Jean-Armand Moroni
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  • "When you prove that "A and B are not in contradiction", what you really prove is that it cannot be proven that (A∧B) is false". Do you mean instead that (A↔B) can be proven true? After all, 1+1=8 and 1+1=9 aren't in contradiction yet their conjunction is false. – ryang May 16 '23 at 03:10
  • @ryang "Being in contradiction" is not a usual concept, at least not that I know. Moreover, "Not being in contradiction", which the OP uses, entails some kind of negation-as-failure, as in Prolog. So for "not being in contradiction" I propose a definition which is "cannot be proven that $(A \land B)$ is false". If the theory is consistent and complete, it implies that $(A \land B)$ is true (except if there are free variables). In your example, it can be proven that $(A \land B)$ is false (both $A$ and $B$ are false), so according to my definition, $A$ and $B$ are in contradiction. – Jean-Armand Moroni May 16 '23 at 12:08
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    @ryang You are right, my answer was unclear. You gave me the opportunity to clarify by answering your first comment, thanks. – Jean-Armand Moroni May 17 '23 at 12:50
  • I'd say that A and B being in contradiction means precisely that A↔B is unsatisfiable (rather than that A∧B is false/unsatisfiable). To wit: 1≠1 and x≠x have an unsatisfiable conjunction yet aren't in contradiction. – ryang Dec 09 '24 at 15:38
  • @ryang You are right, "$A \Leftrightarrow B$ is unsatisfiable" is the best meaning for "$A$ and $B$ being in contradiction". Yet, to "prove by contradiction that $A \Rightarrow B$", we assume $A$ and $\neg B$, and prove it leads to False. So the paragraph beginning by "In proof by contradiction, the situation is different", which is the most useful part of the answer (for OP's question) is still true. I'll change or delete the first part when I find time (which is not the case currently). – Jean-Armand Moroni Dec 10 '24 at 09:56
  • Yes, that mis-definition doesn’t take away from the larger point (explained in my Answer below): even if P & C are both true (this is a stronger condition than merely being consistent with each other, or merely not being in contradiction with each other), P still doesn't generally logically entail C. – ryang Dec 10 '24 at 14:51
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To show that an argument is valid, why can we not

assume that  (I) both its premises and conclusion are true

then show that  (II) there's no contradiction?

(II) is in fact an inevitable result of (I): without introducing any false assumption it is impossible to validly derive any contradiction from a set of true statements.

Moreover, $P_1,\ldots, P_n, C$ all being true in the given context doesn't guarantee that $P_1\land\ldots\land P_n\to C\tag*{}$ is a tautology, which is what it means for the argument $P_1\land\ldots \land P_n;\,\text{therefore, }C\tag*{}$ to be valid (that is, for its premises to logically, even in an alien context, force its conclusion to be true). For instance, the argument 7>4; therefore, every squared number is nonnegative is invalid even though in real analysis its premise and conclusion are both true: in complex analysis, its premise doesn't imply its conclusion.

If $x^2=4$ and $x\neq-2,$ then $x=2.$

Proof

  1. Suppose that $x=2$ (assuming the conclusion).

  2. Then $x^2=4$ and $x\neq-2.$

  3. Thus, if $x^2=4$ and $x\neq-2$, then $x=2.$

Your main argument, which corresponds to the tautology $$\color\red {C}\implies (A \land B\implies C),$$ hasn't actually proven the given statement, because it begs the question and hasn't discharged the supposition $\color\red C.$

With proof by contradiction, it is legitimate to assume that the conclusion is false; why?

Because this assumption is merely provisional and discharged by the end of the proof.

ryang
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  • 7>4⟹ every squared number is nonnegative

    Why is that an invalid argument? If the elements are real numbers, then that should be valid?

    "Definition. An argument is valid if and only if in every case where all the premises are true, the conclusion is true"

     – enoopreuse22 May 17 '23 at 18:26
  • when you say $P_1\wedge ...\wedge P_n$, is each $P$ a separate statement or a valuation of the same statement? e.g $P_1 = (x^2=4)$ $P_2 = (x\neq -2)$, or $P_1 = (1^2=4)$ $P_2 = (2^2=4)$ – enoopreuse22 May 18 '23 at 20:02
  • ok so one way to make step 3 valid would be to make it like this: "if x=2, then if x^2=4 and x≠−2, then x=2.", right? Im not sure if the entire proof would be valid, because the theorem is still not proved.

    Also how could i have discharged the assumption that $C$ is true?

     – enoopreuse22 May 18 '23 at 20:21
  • @enoopreuse22 1. The validity of an argument is independent of context (or any particular world). John is my brother, therefore dogs bark isn't a coherent argument despite corresponding to a true implication. What I call the 'context' is what your definition ("...in every 'case' where all the premises...") calls 'case', and these contexts/cases have a many-to-one correspondence with the truth table rows of the conditional $P_1\land\ldots\land P_n\to C.\quad$ 2. Each $P_i$ just represents a premise, and $C$ the conclusion. – ryang Dec 09 '24 at 16:31
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I have to wonder about your thought process here coming up with this example. Why did you decide to assert that $x\neq -2$? I assume it's because if you said

$$\text{Suppose } x^2 = 4\text{. Then, } x = 2. $$

that you wouldn't be true. But that's exactly the problem with your "proof": it can also be used to prove false claims. After all, if we suppose $x=2$, we see that $x^2 = 2^2 = 4$. There is no hypothesis that rules out $2$, and so we must conclude that the statement is true. It's not though, because $x = -2$ is also not contradicted by the hypothesis.

To boil it down, your "proof" technique is "proof by converse".

To prove $p\implies q$, you show that $q\implies p$.

To prove that $(x^2=4 \wedge x\neq -2) \implies (x = 2)$, you show that $(x=2) \implies (x^2=4 \wedge x\neq -2)$.

If $p$ and $q$ are true, these statements are equivalent. But if $p$ is true and $q$ is false, $q\implies p$ is true while $p\implies q$ is false. This means that your technique will sometimes claim something is true when in reality it is false, thus not a proof. By contrast, the proof by contrapositive works because $p\implies q$ and $\neg q \implies \neg p$ are equivalent. This also implies that a "proof by inverse" would also not work, since it is equivalent to your "proof by converse".

Alex Jones
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