When I was studying, the mathematical analysis professor said something interesting when he was explaining the implication (logical operator), namely $(False \implies True) = True$. He said something like (from my memory):

You can derive any truth from a falsehood. If we accept a single falsehood as truth, then we can prove any single theorem we want.

If I understand it correctly, it means that if we assume that $2+2=5$, then we can provide proof that $\pi = 3$, or that $1 = 2$ or that $\sin^2 x + \cos^2 x \neq 1$.

Is the bold statement true? Is it possible to canonically prove it even though it assumes (temporarily) accepting falsehood as truth?

Just to clarify a little, my question is primarily about the professor’s statement and not about the principles of implication. I’m just curious, what is the scale of destructiveness (in terms of drawing otherwise logical conclusions) of accepting something false as a truth.

It is also interesting to me whether this observation can be generalized, in terms of if we, e.g., falsely assume that dogs and cats are exactly the same animal, can we prove (with a series of otherwise logical conclusions) that, e.g., the Statue of Liberty is actually placed underwater or that the Moon is heavily populated by chipmunks (but that’s a bonus question, because it expands outside the field of mathematics, I guess.)

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    $\begingroup$ Not enough for an answer, but a comment about your real-world dog=cat => Liberty underwater. The difficulty with real-world implications is that it's unclear what "If dog and cat are the same animal" means exactly. We are perfectly capable of imagining a fictional universe in which dogs and cats are the same animal, and the Liberty statue would not necessarily be underwater in that fictional universe. However, in our real universe, yes, the logical implication "If dogs and cats are the same animal then the Liberty statue is underwater" is perfectly true. $\endgroup$
    – Stef
    Commented Dec 6, 2023 at 9:16
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    $\begingroup$ The prime example of this is that if $2+2$ were to equal $5$, then I’m (or you’re, or Bertrand Russell is) the Pope. $\endgroup$
    – Rócherz
    Commented Dec 6, 2023 at 10:24
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    $\begingroup$ @RobbieGoodwin That's pretty much how I look at it as well ... Unless it is an assumption-to-be-rejected-later-on (and in which case anything you prove from that assumption doesn't necessarily hold in that world anyway), 'accepting some statement as false' to me simply says that you are moving to a different world ... and while in that world you can now infer things from that assumption, it's certainly not true that one can 'prove anything you want'. $\endgroup$
    – Bram28
    Commented Dec 6, 2023 at 22:39
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    $\begingroup$ xkcd.com/704 $\endgroup$
    – PM 2Ring
    Commented Dec 7, 2023 at 3:47
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    $\begingroup$ A quite popular group theory exercise states: Prove that a nonabelian group of order $15$ is centerless. Though the premises is false (there isn't any nonabelian group of order $15$), the problem is perfectly legitimate, and solvable. Can we then "prove any single theorem we want"? If so, I wouldn't know how. $\endgroup$
    – Kan't
    Commented Dec 8, 2023 at 22:05

11 Answers 11


Strangely, this is pretty much correct. It is difficult to come up with a reasonable logic that doesn't permit this.

Your arithmetic examples are easy to deal with. If $$2+2=5$$ then we can proceed:

$$\begin{align} 2+2 &= 5 \\ 2 & = 3 &\text{subtract $2$}\\ 0 & = 1 &\text{subtract $2$ again}\\ 0 & = \pi - 3 &\text{multiply by $\pi-3$} \\ 3 & = \pi &\text{add $3$} \end{align}$$

as you requested. (It's similarly easy to get $1=2$ or your other formulas.)

But the problem goes deeper than arithmetic. The logical principle here is called the principle of explosion or sometimes “EFQ” for short, and it is very difficult to avoid.

Suppose our logical system has an “or” operation $\lor$ where $$X\lor Y$$ means $$X \text{ is true, or } Y \text{ is true, or both.}$$ Then these rules both make perfect sense:

  1. If we have proved $X$, we can conclude $X\lor Y$ for any $Y$ at all.
  2. If we have proved $X\lor Y$ and we know that $X$ is false, we can conclude $Y$.

If you accept these, then you should also accept EFQ. This is why: Suppose we have proved $X$, which is false. By (1) we can conclude $X\lor Y$, and since $X$ is false, by (2) we can conclude $Y$. So after proving $X$, which is false, we can conclude $Y$ for any $Y$ at all.

If you don't like this result, you need to say which of (1) and (2) you will reject.

(I asked about this previously. It is hard to get rid of.)

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    $\begingroup$ You seem familiar with paraconsistent logic, and it's referenced in an answer on the linked question you ask. IMO that answers op's original question: "Does one accepted false statement allows proving anything?" -- no, not in paraconsistent logic. It seems like it should be included in this answer. $\endgroup$
    – Burnsba
    Commented Dec 6, 2023 at 16:33
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    $\begingroup$ Paraconsistent logic isn't widely accepted or even widely studied. I tried to suggest that EFQ wasn't absolutely unavoidable, and to point to sources of more complete information, without distracting OP with obscure side excursions. $\endgroup$
    – MJD
    Commented Dec 6, 2023 at 18:17
  • $\begingroup$ I wonder what kind of formalization would allow proofs by contradiction. EFQ alone does not imply that if every statement is true, then something is wrong, whereas human logic, including basically all mathematical research, does that. Simply stating $x\oplus\overline x\equiv\text{True}$ results merely in a contradictory formalization. $\endgroup$
    – rus9384
    Commented Dec 6, 2023 at 21:09

Others have given good reasons why the statement that the professor presumably intended is correct.

However, the actual statement

If we accept a single falsehood as truth, then we can prove any single theorem we want

is not quite correct. It is not enough to accept a falsehood as truth: we would need to accept something that is provably false.

After all, suppose we have a true statement $S$ that we want to prove. One strategy for doing this is as follows: assume $\neg S$, deduce that $0=1$, conclude by contradiction that $S$ is true. The statement above claims that we can always carry out the middle step, i.e. it implies

Any true statement can be proven by contradiction.

But we know that, in any reasonable axiomatic system, there are true statements that can't be proved.

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    $\begingroup$ Also, there are true statements that can maybe be proven, but are hard to prove. If you start with "The Riemann hypothesis is false", you probably won't be able to derive everything and anything from that premise - or if you did, you'd have effectively proven that the Riemann hypothesis is true, which maybe it is but no one has successfully proven it yet. $\endgroup$
    – Stef
    Commented Dec 6, 2023 at 9:18
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    $\begingroup$ @Stef true, but I was interpreting "we can prove" in its weakest possible form of "there exists a proof". $\endgroup$ Commented Dec 6, 2023 at 9:24
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    $\begingroup$ @JustinHilyard the point is that adding $2+2=5$ as an axiom only works because you can prove $2+2\neq 5$ from the other axioms. If the new axiom can't be proved false from the other axioms, there will still be statements that you can't prove using it. $\endgroup$ Commented Dec 6, 2023 at 18:04
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    $\begingroup$ –1; this answer is a nitpicking point of language, irrelevant to this specific question, and just confuses the issue. In ordinary mathematical language, phrasings like “If we accept X, we can prove Y” mean “We can prove Y from the assumption X”. The alternative reading you suggest — “if X holds, then Y is provable [from some fixed axioms, without additional assumptions]” — would only be considered in rather specific contexts, and in those contexts, there’s well-established terminology for avoiding this ambiguity. So the only natural reading of the quote is the intended, correct, one. $\endgroup$ Commented Dec 6, 2023 at 19:39
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    $\begingroup$ @justhalf: No, you're missing a key point. Look at rule 2 in MJD's answer: "If we have proved X∨Y and we know that X is false, we can conclude Y." We need to know that X is false to apply that rule. If we accept a false statement X, but we don't know it's false, we don't have what we need to prove an arbitrary Y. $\endgroup$ Commented Dec 8, 2023 at 0:13

Speaking in algebric terms, you can always prove $a=b$ after doing a false statement.

$\\$$2+2=5\\ 0=1\\ 0=(b-a)\\ a=b$

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    $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – CrSb0001
    Commented Dec 6, 2023 at 0:13
  • $\begingroup$ I agree with previous comment, in particular the third step should state that is obtained after multiplying both sides of the equation by (b-a) $\endgroup$
    – Oliver
    Commented Dec 22, 2023 at 22:39

It is important to be careful in differentiating truth, implication, and proof .. if you find your professor's statements hard to believe (or at least 'weird' enough to spurn you to post a question here), I suspect either your professor was conflating a few things here, or you are misinterpreting what they said.


In mathematics and logic we often use a certain notion of implication, expressed as as 'if ... then ...' statement, where all we mean to say is that it is not the case that the 'if' part is true and the 'then' part is false. I assume your professor was showing you the truth-table for this, and the truth-table shows that that kind of implication is automatically true when the 'if' part is false. Indeed, no matter what the 'then' part says, and no whether that 'then' part is true or false, the whole implication is true. And I suspect that that is all that your professor was trying to say.


A proof is different from an implication. In a proof we say: "This statement is true. Therefore, this other statement is true.". So this is not a statement. but an argument, where you infer one statement (the conclusion) from one or more others (the premises or assumptions). And the argument is an actual proof if the argument is valid, meaning that every step follows agreed upon inference principles such as Modus Ponens or Simplification.

So, given that your professor seemed to merely talk about implications and their truth-conditions, I find your professor's use of the word 'prove' a little out of place.

Still, something similar is going on when it comes to proofs. Again, a proof is valid if the conclusion logically follows from the premises. Another way to think about this is: An argument is valid if and only if in all logically possible worlds where the premises are true, the conclusion is true as well.

So, consider now a statement like 'bananas are pink and bananas are not pink'. This kind of statement is called a contradiction, because it cannot be true: there is no logically possible world where this statement is true. But if there is no world in which it is true, then it becomes vacuously true that in all (zero!) worlds where it is true, something else is true as well. That is, when it comes to proofs, we can derive ('prove') anything from a contradiction.

Proof and Truth

Now, notice how I put 'proof' in parentheses ... because this does seem weird. "Wait! I can prove that the Statue of Liberty is underwater from a contradiction?!? How can I prove something that isn't so?". Well, remember that all you do with a proof is to show that if you assume something to be true, then something else is true as well. But that doesn't mean the latter thing is in fact true ... if the thing you assume is not true, then all bets are off as far as the truth of the latter statement.

So, if you want to know what is really true, you don;t just need a valid argument, but you also need to make sure the assumptions are true in the first place. An argument that has true assumptions is called a well-founded argument, and an argument that is valid and well-founded is called a sound argument. So, with a sound argument you can actually learn something about the world ... a valid proof is not enough to say anything about truth itself.

Contradiction and Falsehood and Proof

I also want you to understand that a contradiction is not the same as a false statement. The statement 'bananas are pink' is false, but it is not a contradiction. Indeed, from 'bananas are pink' alone we really cannot prove that 'the Moon is heavily populated by chipmunks'. Why? It is because we can imagine logically worlds where bananas are pink, but where the Moon is not heavily populated by chipmunks. So, the latter does not logically follow from the former. So you cannot prove 'anything you want' from a false statement.

Mathematical Proof

Now, what about the $2+2=5$ example? Yes, in the world of numbers, it is not true that $2+2=5$ ...but that does not mean that you can logically infer anything from that statement alone. Indeed, note how @MJD's answer derive s $\pi = 3$ with the help of other statements .. statements we normally hold true as well within the world of numbers.

So, we could say that within the world of numbers, if $2+2=5$, then $\pi =3$. As such, we can also say that $\pi =3$ mathematically follows from $2+2=5$, but we can not say that logically $\pi =3$ follows from $2+2=5$. In terms of logic, the best we can say is that $\pi =3$ follows from $2+2=5$ together with the statements we normally regard as true in the world of numbers.

Again, though, you have to understand that proving something does not mean that that thing is true. As we saw, I can mathematically prove that $\pi = 3$ if you assume that $2+2=5$. But of course that doesn't mean that in the world of numbers $\pi =3$ is in fact true.

What does this all Mean?

OK, so now let's generalize all of this, because I think that speaks more directly to your question as to how all of this generalizes. That is: is it true in general that from a false statement you can prove anything you want?

Well, now we know how to unpack this. First, by 'false' we mean 'false in a certain world'. So, consider some statement $A$ that is false in some world. Now, again, the first thing to note is that if $A$ is false in that world, then the statement $A \to B$ is true in that world, no matter whether $B$ is true of false in that world. And that has nothing to do with proofs. OK, but what about proofs? Well, again, we need to make a clear distinction about things that I can prove about that world, and things that I can prove about any world. So yes, if we assume that $A$ is false in a world, then we can prove that for that world, anything else $B$ would be true as well. Of course, as pointed out, that does not mean that $B$ is true ... or that it is false. We know nothing about $B$, because we started with a false assumption to base our proof of $B$ on. Even worse, just because I can prove $B$ for that world, that does not mean that I can prove anything I want in general: there may be other worlds where $A$ is true but $B$ is false.

So, can I prove anything from the assumption that 'bananas' are pink? Well, I certainly cannot prove logically that the Statue of Liberty is underwater from that. I can prove for our world that the Statue of Liberty is underwater, because in our world bananas are not pink, so that contradicts the assumption that bananas are pink, and from a contradiction I can infer anything. But no, this 'proof' does not mean that the Statue of Liberty is actually underwater in our world.

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    $\begingroup$ Don’t you just need the usual axioms of arithmetic, Modus Ponens, and logical axioms sufficient for quantification and the Deduction Theorem, negation and falsum notwithstanding? $\endgroup$
    – PW_246
    Commented Dec 6, 2023 at 0:48
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    $\begingroup$ @PW_246 Yes, that will suffice. But my point was that without any such axioms you cannot derive $\pi = 3$ from $2+2=5$ $\endgroup$
    – Bram28
    Commented Dec 6, 2023 at 2:33
  • $\begingroup$ "Indeed, no matter what the 'then' part says, and no whether that 'then' part is true or false, the whole implication is true. And I suspect that that is all that your professor was trying to say." but the if part being false making the implication true, combined with accepting the if part as true, means that the conclusion must be accepted as true. Which is surely where the professor was going with this. $\endgroup$
    – Cruncher
    Commented Dec 8, 2023 at 23:30
  • $\begingroup$ @Cruncher An ‘ if … then …’ is a statement, not an argument. And the ‘then’ part of an implication is called the consequent, not the conclusion. So it’s different. $\endgroup$
    – Bram28
    Commented Dec 9, 2023 at 1:25
  • $\begingroup$ Bram28 has it right: the implication can be true, but the consequent might NOT be true. $\endgroup$ Commented Dec 13, 2023 at 6:53

It is surprising, and depends on some specifics of the logical system we normally use in mathematics.

If F is false then for any statement A we have the true statement that F implies A (this implication is false only if A is false and F is true, which is not the case).

Then the normal rules of inference tell us that given F (which we are told we are given) and knowing that F implies A, we can infer A.

But A was not specified and can be any statement whatsoever.

So if implication is understood as it usually is, and the rules of inference we are allowed include simple deductions from implications, then from a single false statement we can infer any statement at all which is expressible in the language we are using.


The truth is really about the truth of the conditional statement. There are three components in $p\to q$, they are the statement $p$, the statement $q$, and the conditional statement $p\to q$ as a whole. The truth of the conditional is whether your argument is sound or otherwise. Just because $p$ is false and $q$ is true doesn't make the argument problematic. In fact, the world out there is littered with this kind of valid argument, but which is completely nonsense. Consider the example, "because of the president's incompetency, the economy is in shambles." There is nothing wrong with this argument (the conditional statement $p\to q$). It could have really been the case that the president is doing a poor job (the statement $p$) that results in the economic downturn (the statement $q$), or it could really just be something else (for example, consumer behavior after the pandemic). We all capitalize on this behavior of the conditional statement every now and then to gain an edge against our opponents, big or small, every now and then, knowingly or unknowingly. The scale of destructiveness of this behavior of $p\to q$ is global.

  • $\begingroup$ Great examples! $\endgroup$
    – Oliver
    Commented Dec 22, 2023 at 22:43

The rule of logical implication (modus ponens) has two requirements:

  1. p is true
  2. p-->q is also true.
    Then you can say that q is true.

But p-->q is defined as ~p OR q.

If p is false, then ~p must be true. So p-->q is true and condition 2 above is satisfied. But you CANNOT conclude that q is true because with p false, CONDITION 1 ABOVE IS NOT SATISFIED.

I advise you not to argue this with your instructor; just tell him what he wants to hear, make him happy, and get a good grade. Back in the 1950's I had a great game called WFFNPROOF, where the goal was to construct valid logic formulas using reverse Polish notation. Check it out, if it is still around.


Notice that the statement $~A \implies B~$ is
equivalent to the statement $~(\neg A) \vee B.$

Given statement $~P,~$ and given :

  • Axiom-1 $~P~$ is true

  • Axiom-2 $~\neg P~$ is true.

Then, given any statement $~Q,~$ you have that

$~P \implies (P \vee Q) \iff [(\neg P) \implies Q].$

So, you have

$$\{ ~(\neg P) ~\wedge ~[ ~(\neg P) ~\implies ~Q ~] ~\} ~\implies Q.$$


Intuitively, all true statements are equally true, and all false statements are equally false. There is no such thing as a true statement which is "more true" than another, or a false statement which is "more false" than another. If we accept a false statement as true, that means we have a falsehood which is exactly as true as a true statement. And since all falsehoods are equally true, if we accept even one falsehood as true, any and all falsehoods must be true.

It's simple to construct logical implications which uses two false statements, and we can do so because an implication with a false antecedent is always true - we can always truthfully say that any false statement implies anything else. But if the antecedent is actually taken as true, the second false statement is also implied as true.

"If a dog and a cat are the same thing, then I'm 10 feet tall."

  • $\begingroup$ While this is true for math, it’s not in general true—case in point. “It’s Wednesday” is true today, but isn’t necessarily true. $\endgroup$
    – PW_246
    Commented Dec 6, 2023 at 15:15
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    $\begingroup$ @PW_246 I don't quite see what necessarily true has to do with the question. If I say "if it's Wednesday, then I'm 10 feet tall" when it is, in fact, Wednesday, there is no false statement we are calling true in the first place, but that changes if I say it any other day. We can evaluate the truth of a statement at a particular place or time, why does it matter if that truth value could feasibly be different in another place or time? $\endgroup$ Commented Dec 6, 2023 at 15:55
  • $\begingroup$ if all true statements are equally true, then “It’s Wednesday” is equally true as “$1+1=2$”, which is only true when it’s Wednesday. They aren’t always equally true, so they’re not equally true. $\endgroup$
    – PW_246
    Commented Dec 6, 2023 at 20:34
  • $\begingroup$ @PW_246 "It's Wednesday" said on a non-Wednesday is a false statement. There is no reason that should be equally true as 1+1=2. You're talking about a statement that is sometimes true and sometimes is not. When it's true, it's as true as anything else that's true. You seem to be treating "it's Wednesday" as a true statement which actually is sometimes false. $\endgroup$ Commented Dec 7, 2023 at 13:38

Does one accepted false statement allows proving anything?

Only as an intermediate result in a proof (see line 5 in example below). Note that a proof is not complete until every premise/assumption (lines 1 and 2 here) has been discharged.

Using a form of natural deduction, we can prove, for example, that $P \land \neg P \implies Q$ as follows:

(Screen shot)

enter image description here

It is not $Q$, but rather $P \land \neg P \implies Q$ that has been proven here to be a theorem of propositional logic. It tells us nothing about the truth value of $Q$.

(Text version of proof)

Assuming the falsehood...

1   P & ~P

    Prove: Q
    Suppose to the contrary...

    2   ~Q

    3   P & ~P
        Copy, 1

Discharge the premise on line 2

By contradiction...

4   ~~Q
    Conclusion, 2

We can infer only as an INTERMEDIATE RESULT...

5   Q
    Rem DNeg, 4

We complete the proof by discharging the premise
on line 1 to obtain the theorem...

6   P & ~P => Q
    Conclusion, 1

This is not quite true for all logics currently being actively used by mathematicians. In classical logic, indeed one can prove anything by a reductio ad absurdum argument. Indeed, suppose I want to prove proposition $P$. I will carry out a proof by contradiction. Suppose that $P$ does not hold. But notice that $2+2=5$, a contradiction. Therefore $P$ necessarily holds.

Notice however that what we have actually proved is $\neg \neg P$ (the negation of the negation of $P$). In classical logic this is equivalent to $P$. But in intuitionistic logic it is not. I don't know if intuitionstic logic saves the day or not, but at any rate it does not seem to be immediate that from a contradiction one can prove anything, in intuitionistic logic.

Therefore such a "standard" reductio argument does not produce a proof of $P$ in intuitionistic logic. However, there are other proofs that do (see notes below).

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    $\begingroup$ Unfortunately intuitionistic logic does still admit explosion, for the reason MJD pointed out in their answer. You don't need a proof by contradiction, you can show it directly by the fact that P∧¬P implies Q for all Q: if P is true then P∨Q is true, and if P∨Q is true and ¬P is true then Q is true, both of which steps hold in both classical and intuitionistic logic. Like they said, the principle of explosion is really hard to avoid, and basically comes down to throwing out or replacing one of those two steps in one way or another. $\endgroup$
    – Idran
    Commented Dec 6, 2023 at 14:47
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    $\begingroup$ You can see more about this in this question: math.stackexchange.com/questions/2922457/… $\endgroup$
    – Idran
    Commented Dec 6, 2023 at 15:11
  • $\begingroup$ @JustinHilyard, there is a variety of intuitionistic logics. Do you have a source for the claim that they all accept the principle you mentioned? $\endgroup$ Commented Dec 7, 2023 at 10:50
  • $\begingroup$ @JustinHilyard, the question you link does not deal with the same formula as here. $\endgroup$ Commented Dec 7, 2023 at 10:52
  • $\begingroup$ These are fundamental, definitional properties of the disjunction operator; I don't have a specific source for all forms of intuitionistic logic using disjunction, but any logic that uses the standard disjunction and negation operators (with or without double-negation) will therefore admit the principle of explosion. Also, unless I'm understanding incorrectly, you're mistaken; the OP's question is essentially referring to the principle of explosion, and the question linked is about the principle of explosion in intuitionistic (also often known as constructive) logic. $\endgroup$
    – Idran
    Commented Dec 7, 2023 at 14:45

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