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 $x=2.$


    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?

  • 3
    $\begingroup$ That way you could also prove "Suppose $x^2 \ge 4$ and $x\neq-2$. Then $x=2$". $\endgroup$
    – dxiv
    May 12, 2023 at 19:06
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    $\begingroup$ 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 $5\pi/4$.) $\endgroup$ May 12, 2023 at 19:11
  • 2
    $\begingroup$ You've proved that if $x = 2$, then $x = 2$. That's probably not what you're going for. $\endgroup$
    – anomaly
    May 12, 2023 at 19:58
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    $\begingroup$ 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). $\endgroup$
    – David K
    May 13, 2023 at 4:21

4 Answers 4


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.”


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 $A \Rightarrow B$.

  • $\begingroup$ "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. $\endgroup$
    – ryang
    May 16, 2023 at 3:10
  • $\begingroup$ @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. $\endgroup$ May 16, 2023 at 12:08
  • $\begingroup$ I was pointing out that "When you prove that X, what you really prove is that Y" sounds like you are saying that X implies Y, but in fact it is Y that implies X. However, my observation is premised on the idea that "P and Q contradict each other" precisely when they negate each other (since in logic, a contradiction is the conjunction of a sentence and its negation). $\quad$ You're now clarifying that that quotation is a definition; this is fine, but the way you wrote it doesn't read as a definition. $\endgroup$
    – ryang
    May 16, 2023 at 12:54
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    $\begingroup$ @ryang You are right, my answer was unclear. You gave me the opportunity to clarify by answering your first comment, thanks. $\endgroup$ May 17, 2023 at 12:50

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


  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 $x=2.$

In your first two steps, you proved the given statement's converse $$C\implies A \land B.$$ Then I think you implicitly inferred that $$\color\red {C\implies} C \land A\land B,$$ though you didn't make it explicit. Then you implicitly inferred that $$\color\red {C\implies} \Big(A \land B\implies C\Big).\tag0$$ So far, so good. Unfortunately, you carelessly wrote statement $(0)$ just as $$A \land B\implies C,\tag3$$ making Step 3 an invalid inference! You'd neglected to discharge the supposition $\color\red C$ that was made in Step 1.

In other words, you have not validly deduced the required statement $(3);$ you have validly deduced statement $(0),$ which says that if $\color\red C$ (i.e., the supposition $\color\red {x=2}$ that you made in Step 1) is actually true then statement $(3)$ is true. Unfortunately, statement $(0)$ is not useful information, because it is a tautology: no knowledge of mathematics or elementary algebra is required to validly deduce statement $(0),$ which can be proven simply using a truth table.

Begging the question is the fallacy of arguing by assuming the conclusion. In your Step 1, assuming the consequent x = 2 of the given statement amounts to assuming that the statement is true (notice that if B is true then A→B is automatically true), that is, assuming the conclusion of the proof/argument; tautology $(0)$ exhibits the structure of this fallacy.

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?

If the argument's premise $P$ and conclusion $C$ are both true, then by definition they don't contradict each other (i.e., $P\not\equiv\lnot C$).

When you say "show that there's no contradiction" (if you in fact mean "...leads to no contradiction" then how would you actually do that??), do you perhaps instead mean the weaker condition $\text“P_1,\ldots,P_n,C$ are consistent with one another”, which is an immediate consequence of $(P_1\land\ldots \land P_n \land C)$ being true? In other words, are you perhaps just meaning to ask this simpler question:

  • does its premises and conclusion being consistent with one another imply that the argument is valid?
  • does showing that $(P_1\land\ldots \land P_n \land C)$ is true (this implies that $P_1,\ldots,P_n,C$ are consistent with one another) prove that $(P_1\land\ldots \land P_n \to C)$ is a tautology?

The answer is No: in real analysis,

  • $7>4;$ therefore, every squared number is nonnegative

has premise and conclusion both true yet is an invalid argument, because the corresponding conditional is not a tautology, because in complex analysis this conditional is actually false.

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

Because this assumption is merely provisional and is discharged by the end of the proof. On the other hand, your proof attempt does not eventually discharge the $\color\red{\text{assumption}}$ that $\color\red C$ is true. (The point is that when we make a supposition/assumption, we need to keep track of it and remember to also consider the case in which it is actually false.)

P.S. To say that the argument $(P_1\land\ldots \land P_n,\,\therefore C)$ is valid is to say that its premise forces the conclusion to be true even in an alien world/context; remember, we are investigating the logical truth of $(P_1\land\ldots \land P_n \to C)$ regardless of the truth value of "every squared number is nonnegative" in any particular world.

  • $\begingroup$ 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" $\endgroup$ May 17, 2023 at 18:26
  • $\begingroup$ @enoopreuse22 The point is that the validity of an argument is about reasoning, not truth per se, and is independent of context. John is my brother, therefore dogs bark corresponds to a true implication but not a valid argument/reasoning. In the bullet points, I indicate that the argument P, ∴C is valid precisely when P→C is a tautology; referring to your definition ("...in every 'case' where all the premises..."), P→C being a tautology means that as you vary the meanings of P and C (i.e., in every 'case'/context) P→C remains true. My postscript is explaining the same thing. $\endgroup$
    – ryang
    May 18, 2023 at 4:58
  • $\begingroup$ @enoopreuse22 In that definition that you supplied, 'every case' refers to 'every row of the argument's corresponding truth table'. There are infinitely many possible 'cases', but every 'case' belongs to some row of the argument's truth table. $\endgroup$
    – ryang
    May 18, 2023 at 5:01
  • $\begingroup$ 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 $P_1 = (1^2=4)$ $P_2 = (2^2=4)$ $\endgroup$ May 18, 2023 at 20:02
  • $\begingroup$ @enoopreuse22 Each $P_i$ is a different premise. $\quad$ I don't think you are using the word 'valuation' correctly: it means to assign truth values, not variable values. $\endgroup$
    – ryang
    May 18, 2023 at 20:08

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".


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