Consider a statement If $A = B$ then $C = D.$ If one were to prove it, the first thing that he/she might do is to use the fact that $A = B$ and, with some algebraic manipulation, reach the conclusion $C = D$.

Now, I proved it by using some other route.

I assumed that $C = D$ and used it to reach the conclusion that $A = B.$ Since the fact that $A = B$ is true, our assumption was right and thus it is proved that $C$ is actually equal to $D$.

My question is: Is this way of proving something correct?

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    $\begingroup$ No. What you're trying to prove here is "if $Q$ then $P$", which a totally different statement than "if $P$ then $Q$" $\endgroup$
    – Hermès
    Oct 25 '16 at 11:25
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    $\begingroup$ e.g. "If you are a man, then you are human" cannot be used to infer, "if you are human, then you are a man" $\endgroup$ Oct 25 '16 at 12:57
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    $\begingroup$ Something that is valid is to prove that "if $C \neq D$ then $A \neq B$". This is logically equivalent to "if $A = B$ then $C=D$" by contraposition $\endgroup$ Oct 25 '16 at 13:40
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    $\begingroup$ Addtionally, proving "$A=B \Rightarrow C=D$" does not require assuming $A=B$ $\endgroup$ Oct 25 '16 at 16:16
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    $\begingroup$ If you phrase your question as "Can we prove $p\to q$ by assuming $q$ and proving $p$", the answer should be much clearer. You're letting the structure of $p$ and $q$ confuse you. Incidentally, the title of this thread should be revised into something more descriptive. $\endgroup$
    – Asaf Karagila
    Oct 25 '16 at 17:13

No, what you are describing is the logical fallacy of affirming the consequent.

If you have $P\implies Q$, that doesn't mean that $P$ is true just because $Q$ is. Example in propositional logic

It rains, therefore the road is wet.

This is perfectly valid, however just because you see the road wet doesn't mean it has rained. After all I can grab my hose and spray it all over the road for an hour without it having rained or be raining.


With your "route" we prove that 1=0 as follows:

1=0, hence 0=1.

We have:



Adding these equations, we get 1=1, which is true ....




Let us take an example -

If $x = 0$, then $\sin x = 0$. But if $\sin x = 0$, then $x$ is not necessarily 0.

If we start by your method to prove $\sin x = 0 \implies x = 0$, we would end up proving it despite it being a false statement.


It only works if the operations you used to get from $C = D$ to $A = B$ are reversible. For example, $x=2 \to 2x=4 \to 2x+1=5$ does demonstrate that if $2x+1=5$ then $x=2$, but only because every step I took works just as well backwards as forwards. The problem is, almost anything remotely interesting isn't reversible. For example, say we start with $x=2 \to x^2=2x \to x^2-2x = 0$ and conclude that if $x^2 - 2x = 0$, then $x = 2$. That's not true - if $x^2-2x=0$, then $x$ could just be zero. The problem is that the step $x=2 \to x^2=2x$, while perfectly valid, can't be reversed if $x = 0$.

Your approach is a good way to find a proof, but it doesn't make one. What you can do is start with $C = D$, prove that $A = B$, and then try to reverse all of your steps. If every single step is reversible, then the backwards version of your proof is a proof that if $A = B$ then $C = D$. But if reversing doesn't work, you have nothing.

  • $\begingroup$ The exact question is as follows: if $2b^2 = a^2 + c^2$ then prove that $$\frac{2}{b + c} = \frac{1}{c + a} + \frac{1}{a + b}$$. So, does it mean that I can use my way of proving in this particular question? $\endgroup$
    – Parth
    Oct 26 '16 at 15:49
  • $\begingroup$ No, you're not understanding - whether or not it's reversible depends on the sequence of steps you use, and has nothing whatsoever to do with what you're trying to prove. For example, $x=2 \to x + x = 2 + x \to 2x = 2 + x$ is reversible; $x=2 \to x^2=2x \to x^2+x^2=2x+x^2 \to 2x^2=2x+x^2 \to 2x=2+x$ is not. One approach would be to try to use your method on this problem, then see if the steps are reversible. If so, the reversed steps are your proof. If not, you have to try something else. $\endgroup$ Oct 27 '16 at 1:01
  • $\begingroup$ I liked this answer. (It mentions several things which students learning this should be told and keep in mind.) $\endgroup$ Dec 10 '16 at 16:33

This is not valid. You assumed $C=D$, based on that assumption derived $A=B$, and then incorrectly concluded that $A=B$ will imply $C=D$.

What you have shown is that $A=B$ is a neccessary condition for $C=D$: whenever $C=D$ holds, then $A=B$ must neccessarily hold as well. But the usual implication is used to characterize a sufficient condition. The implication would mean that $A=B$ would be sufficient to obtain $C=D$.

To put it differently, you showed that having $C=D$ is one way to obtain $A=B$. But you have not shown that it is the only way. There might be another way to get $A=B$ despite $C\neq D$. Which means $A=B$ tells you nothing about whether $C=D$ unless you can also demonstrate that there can be no other way to reach $A=B$ without $C=D$.

Which is the foundation for a usual proof by contradiction. Assume $A=B$ and $C\neq D$, then find a contradiction. That shows that the case $A=B$ and $C\neq D$ is not possible, so $A=B$ must indeed imply $C=D$.

One example: let $A=x^2, B=9, C=x, D=3$. So you are talking about whether or not $x^2=9$ implies $x=3$. Assuming $x=3$ you can easily show $x^2=9$. But there is another way to obtain $x^2=9$ without having $x=3$, namely using $x=-3$.


The error in your thinking is "use the fact that A=B". Nowhere is it really a "fact" that A=B. You can start with that as an assumption, to try to infer other things that might be based on that assumption (like C=D), but nothing you have said can be used to demonstrate A=B absent that assumption.

If you assume A=B, then you can easily and trivially find a way to prove A=B from that, obviously. But that would be like me saying, I'll assume I'm a genius, therefore that proves I am. Clearly that argument is invalid.

  • $\begingroup$ I think the OP's desire is not to "use the fact that $A=B$", but rather to first assume $C=D$. And the proposition to be proved is neither $A=B$ nor $C=D$ but rather $A=B \implies C=D$. $\endgroup$
    – David K
    Oct 25 '16 at 13:13
  • $\begingroup$ @David C=D does entail (can be used to show that) A=B => C=D, so if he's trying to prove that, that's possible, but I read it as he was trying to prove something in the relationship between A and B, or between C and D, which there is insufficient information to do. $\endgroup$ Oct 25 '16 at 13:22
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    $\begingroup$ Good point. OP wrote "if $A = B$ then $C = D$", then "prove it", in which most of us interpreted "it" to be the preceding implication. But $C=D$ does entail $A = B\implies C = D$, so what he's trying to do could be considered as an indirect application of the argument you described in your last paragraph (which, as you pointed out, is invalid). $\endgroup$
    – David K
    Oct 25 '16 at 13:40

I see some great answers above. But nobody mentions the words sufficient and necessary to my dismay.

$p\to q$ means that p is sufficient for q

You started with $q$ and proved $p$. In other words, you proved p is necessary for q.

They are not the same thing.

In light of the example given by Aortiz:-

If it rains the road is wet.

Rain is sufficient to make the road wet, but not necessary, because there are more ways for road to get wet (Insert cheap humour here).

You may sometimes get lucky when the converse of the statement is also true, like:-

If Barca wins, Madrid loses.

Since exactly one side in football wins, victory of Barca is both necessary and sufficient for defeat of Madrid.

(No offence meant to Madrid fans !)


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