Prenex Normal Form of a Simple Proposition Reads Strangely.

Question

I was trying to convert a simple (true) proposition concerning the real numbers to Prenex Normal Form but arrived at a logical statement that didn't appear equivalent to what I started with.

The proposition I began with:

If two real numbers $a,b$ can get arbitrarily close to one another, then they are equal.

As a logical formula:

if $|a-b|<ε$ for all $ε>0$, then $a=b$.

Entirely with logical symbols:

$ (∀ε>0, (|a-b|<ε)) ⇒ (a=b) $

This logical statement appears identical in meaning to the original proposition above.

Now I perform the following steps to convert this to Prenex Normal Form (PNF) by applying the instructions in https://www.csd.uwo.ca/~lkari/prenex.pdf.

1. Eliminate occurrences of conditionals and biconditionals from the formula. This means removing the $⇒$ symbol in this situation.

   Applying the fact $(p⇒q) ⇔ (¬p∨q)$ gives:

   $ ¬(∀ε>0, (|a-b|<ε)) ∨ (a=b) $

   Observe that the conditional is removed.

2. Move all negations inwards so that negations only appear as part of literals.

   There's only one negation to move and moving it inwards gives:

   $ (∃ε>0, ¬(|a-b|<ε)) ∨ (a=b) $

3. The PNF can now be obtained by moving all quantifiers to the front.

   So we just have to shift the existential quantifier outside and we're done:

   $ ∃ε>0, ( ¬(|a-b|<ε) ∨ (a=b) ) $

   Observe that the formula that comes after the existential quantifier can be rewritten as a conditional since it's in the "$(¬p∨q)$" form. Rewriting the formula as a conditional gives:

   $ ∃ε>0, ( (|a-b|<ε) ⇒ (a=b) ) $

In English, this may be read as:

"There exists a positive real number, call it ε, such that if the absolute difference between and is at most ε, then and are equal."

However, this result does not appear equivalent to the original proposition I began with:

If two real numbers , can get arbitrarily close to one another, then they are equal.

Following @mjqxxxx's suggestion to try making $∀,∈$ and $ε$'s dependency on $a$ and $b$ explicit in the PNF formula gives:

"For all real numbers ,, there exists a positive ε that possibly depends on , , such that if the absolute difference between and is at most ε, then and are equal."

However, this description still reads as being distinct from the original proposition. :(

---

Update: Following the links @NaïmFavier provided, I was able to proof the PNF formula with a strategy used in Smullyan's proof of the Drinking Principle.

Proof ($\quad$$∃ε>0 ( (|−|<ε) ⇒ (=) )$ is True.$\quad$):

There are two cases to consider: either $a=b$, or $a$ and $b$ differ by some amount – i.e., $a=b+ε_0$.

Case 1 ($a=b$): The consequent of the implication is true. By the fact that a true proposition is implied by any proposition, the implication we want to prove holds regardless of whether the antecedent "$|a-b|≤ε$" is true/false.

Case 2 ($a$ and $b$ differ): Suppose that $a=b+ε_0$ for some $ε_0 ∈ \mathbf{R}$. Letting $ε := ε_0$ makes the antecedent "$|a-b|<ε_0$" false. By the fact that a false proposition implies any proposition, the implication holds.

$\square$

---

Related phrases: drinker's paradox, drinker paradox, drinker's theorem, Smullyan's Drinker's principle, Raymond Smullyan's "What Is the Name of this Book?"

Answer 1

It is equivalent, and your confusion is called the drinker's paradox (where "$x$ drinks" means "$|a - b| < x$").

If $a = b$, then you can pick any $\epsilon > 0$ and the statement will hold because $a = b$ is true. If $a \neq b$, then $|a - b| \neq 0$, hence you can pick $\epsilon = |a - b|$ and the statement will hold because $|a - b| < \epsilon$ is false.

Amusingly, I recently saw something very similar to this question formulated as a puzzle on Tim Gower's blog: https://gowers.wordpress.com/2013/12/09/a-little-paradox/

Answer 2

It's equivalent to the original statement. The reason it reads strangely (which is somewhat subjective, but I agree that it does) may be that it's not immediately apparent that $\varepsilon$ can (must) depend on $a$ and $b$. On a casual reading, the sentence seems to be mainly talking about some inherently special value of $\varepsilon$. Putting an explicit $\forall a \forall b$ in front seems (again, subjectively) to more clearly indicate that the statement is made in the context of a particular $a$ and $b$, and much of the strangeness goes away.