# What is wrong with this sort of proof?

Consider the following implication, where a, b and x are real numbers:

(x – a > b → x > a) → b > 0

Key: "ϕ → ψ" just means "ϕ implies ψ".

I will assume that it is self-evidently false.

So, what is wrong with the following proof by counterexample?

The implication (x – a > b → x > a) → b > 0 is false because if a = 33, b = -1 and x = 37, then 37 – 33 > -1 → 37 > 33 is true so x – a > b → x > a is true, and -1 > 0 is false so b > 0 is false.

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Philosophy Meta, or in Philosophy Chat. Comments continuing discussion may be removed. Apr 9 at 8:14
• Proofs prove things. What are you trying to prove? Apr 9 at 10:22

So, what is wrong with the following proof by counterexample?

The implication (x – a > b → x > a) → b > 0 is false because if a = 33, b = -1 and x = 37, then 37 – 33 > -1 → 37 > 33 is true so x – a > b → x > a is true, and -1 > 0 is false so b > 0 is false.

There is nothing wrong with your proof. A more complete counterexample that invalidates both implications is possible.If negative numbers are allowed then a situation exists where both x > a and b > 0 are both false.

The first equation can be expressed as:

x - (a+b) > 0

10 - (12 + -3) > 0 here b < 0 and a > x. So your premise that x>a is also false.

Note that your statement is true if x,a, and b are positive real numbers.

What are you even trying to do here in the first place and how do you define "implies"? Because neither your statement nor your proof make a whole lot of sense.

Now obviously if by "implies" you mean that for all sets of variables for which the left side of the implication is true, the right side MUST also be true, then already the first part is false:

``````x – a > b → x > a
``````

as there are obviously set of a,x,b such as (6,5,-2) for which the left hand side is true, yet the right hand side is false. Now if I understand Mauro ALLEGRANZA correctly you could of course make the argument that `(TRUE → FALSE) → ANYTHING` is a true implication, but obviously you couldn't just follow from that that ANYTHING is actually true. For that you'd need to show that the argument is indeed sound and that both TRUE and FALSE are true, which they aren't (FALSE is false...).

So what is it that you're trying to proof? That `b > 0` is false or that the implication is false?

I mean as said the implication might be true, even though that is not really any useful (at least from what I can see). And no you obviously can NOT deduce that b <= 0 as you did here:

The implication (x – a > b → x > a) → b > 0 is false because if a = 33, b = -1 and x = 37, then 37 – 33 > -1 → 37 > 33 is true so x – a > b → x > a is true, and -1 > 0 is false so b > 0 is false.

That's where I guess Conifold's comment becomes relevant in that it makes a huge difference as to whether it's `∀x∀a∀b[(x – a > b → x > a) → b > 0]` or `∀b[∀x∀a[(x – a > b → x > a)] → b > 0]`. For example if I were to chose any b>0 then for all a and for all x for which `x-a>b`it would follow that x must be bigger than a and it would further follow that b is bigger than 0.(Obviously that would be circular reasoning as I imply b>0 from an assumption of b>0, so not really useful, but still obviously true).

So no it's a non sequitur to assume that b>0 is false, there is an infinite amount of possible configurations of a,b,x for which b is indeed bigger than 0.

I mean let's drop the math essentially the statement is "total nonsense"-> b>0 Now just because the implication doesn't work and you can't follow b>0 from total nonsense doesn't mean that b cannot still be >0.

And obviously as the left hand statement could be true or false, depending on the choice of parameters, you cannot claim that it is self-evidently false.

This is really just expanding on points made by Conifold and PW_246 in the comments.

(x – a > b → x > a) → b > 0

This is called an open formula, because the variables x, a and b are not bound by any quantifier. It would be a little like writing in English, "___ is red". What does it mean? Some thing is red? Everything is red? We can’t tell. Without some context, it doesn’t make sense. Or how about, "___ is greater than ___". Without some other information, we don’t know what is being stated.

Similarly, the variables x, a and b in the given formula are like placeholders where some value has to be filled in. But without some context or some convention, we don’t know whether the formula is stating that it holds for all possible values of x, a, and b, or for some range that depends on the relation between the variables.

A default convention that is often employed is to suppose that unbound variables are implicitly universally quantified. So, for example, Fx might be treated as being equivalent to (∀x)Fx. The convention is not universal, so context is still needed.

Now if we suppose that your formula is implicitly universally quantified, then we would have:

(∀x)(∀a)(∀b)((x – a > b → x > a) → b > 0)

That is indeed false for the real numbers. We only need one interpretation under which it is false to serve as a counterexample, and your example of a = 33, b = –1 and x = 37 fits the bill.

You don’t say where the formula comes from, but it may be that the author is trying to express the fact that if we are able to deduce x > a from x – a > b for any x and a, then it follows that b > 0. That is true. Consider that x > a is equivalent to x – a > 0. So if from x – a > b it follows that x – a > 0, then b > 0.

This can be expressed in a sentence as:

(∀b) [(∀x)(∀a)(x – a > b → x > a) → b > 0]

This is true because the antecedent part of the conditional: (∀x)(∀a)(x – a > b → x > a) is false when b is negative or zero.

A big part of the problem here lies with interpreting the word ‘implies’, which is ambiguous in English. Unfortunately, mathematicians get very sloppy with this word. They are usually taught to call the material conditional material implication and to read P → Q as “P implies Q”. This is really rather misleading. It blurs the important distinction between a conditional in the object language and a consequence relation in the metalanguage. It is important not to fall into this trap. If you’d like, ask a separate question on this topic and I’ll explain it in more detail.