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?

  • $\begingroup$ 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. $\endgroup$ – AnyAD Jun 9 at 22:58
  • $\begingroup$ Is the inverse $B $ unique in your case ($B=A^{-1}\iff AB=I$)? $\endgroup$ – AnyAD Jun 9 at 23:08
  • $\begingroup$ @AnyAD Yes but I haven't proven so yet. $\endgroup$ – 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 ($B=\dots$))

  • $\begingroup$ 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. $\endgroup$ – holomenicus Jun 10 at 1:33
  • $\begingroup$ 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. $\endgroup$ – 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.


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