# How to write an implication whose antecedent quantifies a variable in its consequent?

I want to write the statement

$$\text{If } A^{-1} \text{ exists then } A^{-1} = \frac{\alpha-a}{\alpha^2-a^2}$$

using quantifiers. Note that $$A$$ and its inverse, if it exists, are taken from the set $$\mathscr{G}$$. My first thought is

$$\exists B \in \mathscr{G}, BA = 1 \to B = \frac{\alpha - a}{\alpha^2 - a^2}$$

but this is incorrect because the antecedent needs to include the quantifier while here the grouping works out like so:

$$\exists B \in \mathscr{G} \left( BA = 1 \to B = \frac{\alpha - a}{\alpha^2 - a^2} \right).$$

This is still true if $$A^{-1}$$ does not exist; in this case, there is simply no $$B$$ that satisfies the antecedent. But this is not the statement I'm looking for, because I need the quantifier in the antecedent. But the difficulty is that with the quantifier in the antecedent, there's no way to "instantiate" $$B$$ in the consequent:

$$\left( \exists B \in \mathscr{G} , BA = 1 \right) \to \left( \underbrace{B}_\text{Not the same B.} = \frac{\alpha - a}{\alpha^2 - a^2} \right).$$

I believe that in first order logic you can say, "if there exists some $$x$$ satisfying these properties, then let this variable here be that $$x$$." How do I write that in this context?

• The statement is about any $A$ in the set. So quantify $A$ first. Then if a $B$ exists, we have that $B=\dots$. And this is done. Everything is quantified and implication is clear. – AnyAD Jun 9 at 22:58
• Is the inverse $B$ unique in your case ($B=A^{-1}\iff AB=I$)? – AnyAD Jun 9 at 23:08
• @AnyAD Yes but I haven't proven so yet. – holomenicus Jun 10 at 19:54

Your statement is what is called a 'donkey-sentence', named after the rather violent example of:

'If a farmer owns a donkey, he will beat it'

Note how if you are asked to symbolize this one, you are likely to go through the same thought process you went through:

The first thought is that it is a conditional:

$$\exists x \ \exists y (F(x) \land D(y) \land O(x,y)) \to B(x,y)$$

But this doesn't work, since you have free variables in the consequent.

OK, but if you then pull the scope of the existential over the consequent, you get:

$$\exists x \ \exists y ((F(x) \land D(y) \land O(x,y)) \to B(x,y))$$

which doesn't work either, since this statement can be made trivially true by picking anything for $$x$$ that is not a farmer and/or anything for $$y$$ that is not a donkey, and hence the statement ends up saying little of interest at all.

The correct answer, of course, is to use a universal:

$$\forall x \ \forall y ((F(x) \land D(y) \land O(x,y)) \to B(x,y))$$

because the 'a' in 'a farmer' and 'a donkey' is really used as 'any'.

Similarly, the correct symbolization of our statement should be:

$$\forall B \in \mathscr{G} \left( BA = 1 \to B = \frac{\alpha - a}{\alpha^2 - a^2} \right).$$

Now, you may still be wondering why you have a feeling that there should be an existential here.

Well, consider the following modified donkey-sentence:

'If a farmer owns a donkey, then we have a problem'

Now this sentence can be symbolized as:

$$\exists x \ \exists y (F(x) \land D(y) \land O(x,y)) \to P$$

Moreover, this turns out to be equivalent to:

$$\forall x \ \forall y ((F(x) \land D(y) \land O(x,y)) \to P)$$

So yes, there is a close connection between the use of the existential and the universal when conditionals are involved.

$$\forall A,B \in S ( (AB=I)\implies B=\dots )$$

Or you can write: for all $$A$$ in the set (( there is B in the set AB=I) therefore ($$B=\dots$$))

• This proposition is just the same as the third one in my question. I need the quantifier in the antecedent, not quantifying the whole implication. – holomenicus Jun 10 at 1:33
• What about $\exists B$ as you wrote? What is $B$ in the consequent (do you need to quantify $\alpha$,$a$)? I am affraid I don't really understand your question. – AnyAD Jun 10 at 2:23

You can actually just write $$A^{-1} = B \implies B = \frac{\alpha - a}{\alpha^2 - a^2}$$

since the equivalence will only be provable if $$A^{-1}$$ is defined.