Here's what we want to prove: $\forall$x$\in$$\mathbb{R}$$(x=0)$ $\vee$ $\forall$x$\in$$\mathbb{R}$$(x$ $\ne$ $0)$ .

We know that $\forall$x$\in$$\mathbb{R}$ $(x=0$ $\vee$ $x$ $\ne$ $0)$.

Now, let x be arbitrary. We know that either $x=0$ or $x$ $\ne$ $0$. Let's assume $x=0$. Then, since $x$ was arbitrary, we have $\forall$x$\in$$\mathbb{R}$$(x=0)$.

If $x$ $\ne$ $0$, then again, since it was arbitrary, we have $\forall$x$\in$$\mathbb{R}$$(x$ $\ne$ $0)$.

Thus, in either case we have $\forall$x$\in$$\mathbb{R}$$(x=0)$ $\vee$ $\forall$x$\in$$\mathbb{R}$$(x$ $\ne$ $0)$.

well, I know that the error is in "Then, since $x$ was arbitrary, we have..." i just don't know how to explain to myself why this is logically incorrect.

  • 1
    $\begingroup$ "let $x$ be arbitrary. Let's assume $x=0$." Now $x$ is no more arbitrary and thus we cannot "generalize" to get $\forall x \ (x=0)$. $\endgroup$ – Mauro ALLEGRANZA May 22 '17 at 10:55

If you intend to prove a statement of the form $\forall x \in M : P(x)$, where $P$ is some statement, then you may prove it in the following way:

Let $x \in M$ be arbitrary. Then some arguments follow...and therefore we assert $P(x)$. Since x was arbitrary, the statement follows.

What you do in the proof is different. You prove only one part of $P(x)$ and then take a new $x$. Your Since $x$ was arbitrary comes too early, since you have only treated the first case.

The right way would be to say: "Take an arbitrary $x$, then we have two cases. Case 1: $x=0$. Case 2: $x\neq 0$". And you first have to get through with both cases before saying that it now follows for every $x$. Then you see that you can't prove the obviously wrong statement.

  • $\begingroup$ so, with a given in the form of disjunction it must be used as a whole inside of $[$let $x$ be arbitraty $-$ since $x$ was arbitrary$]$ construction? $\endgroup$ – famesyasd May 22 '17 at 10:42
  • $\begingroup$ Yes, I think you can say it like that. $\endgroup$ – Luke May 22 '17 at 12:44

You are assuming that $x=0$ for arbitrary $x\in\mathbb{R}$ which is the same as for each $x\in\mathbb{R}, x=0$. Your argument is circular. You are assuming your conclusion. And note that for any proposition $p$, $p\implies p$ is true.


Along the lines of @Luke (already accepted) ...

Now, let x be arbitrary.

I think you mean, let $x$ be an arbitrary element of $\mathbb{R}$. It makes a difference.

We know that either $x=0$ or $x$ $\ne$ $0$.


Let's assume $x=0$.Then, since $x$ was arbitrary, we have $\forall$x$\in$$\mathbb{R}$$(x=0)$.

This would not follow even if we assumed $x\in \mathbb{R}$. If we assume that something is true of one element of a set, it does not immediately follow that it is true of every element of that set. If, for example, we assume that $x$ is a human female, it does not follow that every human is a female.


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