1

Suppose that I have the axiom: $ \forall x\colon x=x. $

Which formal rule can be used to derive: $\forall x \exists y\colon x=y$.

I need to apply existence introduction, but not at the top level, but inside a quantified predicate...

added clarification by the answer and comments I understand that my problem is something more basilar. In my understanding of a formal system every theorem is obtained from the application of a inference rule applied to previously obtained theorems or axioms . The proposed answer instead assumes that one can make temporary assumption to draw a conclusion. I understand that this is how mathematicians usually write proofs, I wonder if this is how the formal system is actually defined.

Emanuele Paolini
  • 22,203
  • 2
  • 40
  • 70
  • Apply $\exists$-introduction, then $\forall$-introduction. In $\lambda$-calculus: $\lambda x. (x, \mathrm{refl}(x))$. – Naïm Camille Favier Jul 18 '24 at 19:30
  • 1
    If you need to introduce operators "inside" a formula this means that you first need to apply rules to eliminate the outer operators (and then later reintroduce them after you intruduced the one in the middle). – Natalie Clarius Jul 18 '24 at 20:05
  • Making temporary assumptions is allowed in some formal systems, but the given answer doesn't do that; it's a straightforward application of the standard inference rules of natural deduction for first-order logic, which it sounds like you need to carefully review. – Naïm Camille Favier Jul 19 '24 at 09:17
  • A more "mathematical" version of this problem might be to prove $\forall a\in D: \exists b \in D: a=b$ where $D$ is the domain of discussion. If you are allowed this, the proof would be more straightforward. The first line would be to assume $x\in D$. – Dan Christensen Jul 19 '24 at 21:13
  • @NaïmFavier in the given answer, the second line, a=a, is a theorem? I would say it is an assumption. – Emanuele Paolini Jul 22 '24 at 04:46
  • 1
    @EmanuelePaolini It is a theorem, in a context with one free variable $a$. The only assumption is $\forall x. x = x$, because that is your axiom. – Naïm Camille Favier Jul 22 '24 at 08:52
  • @NaïmFavier this is the point I'm trying to understand. In the proof below it is tacitly understood that the second and third line are inside a context (maybe one could indent them to make this apparent) and that the variable a only exists in this context. This means that inference rules should mention the possibility to enter and exit a context. In the wikipedia page that you have mentioned this is maybe hidden in the use of the ⊢ symbol which seems to me that is not clearly defined. Maybe I will ask a new question now that I have more clearly understood which is my problem. – Emanuele Paolini Jul 23 '24 at 12:23

1 Answers1

2

Just at each stage do the obvious thing!

$\forall x\ x = x\quad\quad$ premiss
$a = a\quad\quad\quad\ \ \forall$E
$\exists y\ a = y\quad\quad\ \exists$I
$\forall x\exists y\ x = y\quad\ \forall$I

The only sensible way to use the premiss is to instantiate the universal quantifier with a constant/parameter/arbitrary name, which gets us the second line. What else?!

Existential quantifier introduction gives us the third line. Then, since $a$ doesn't appear in any premiss we can universally quantify to get what you want.

Peter Smith
  • 56,527
  • I'm bothered by the second and third lines, which contain a predicate instead of a proposition. Isn't it valid to deduce the conclusion directly from the premise? – Emanuele Paolini Jul 19 '24 at 08:24
  • 1
    @EmanuelePaolini Using which rule? – Naïm Camille Favier Jul 19 '24 at 09:18
  • In your 2nd line, you introduce a free variable by universal specification/instantiation. I know that this is accepted practice in standard FOL, but readers should know that it is NOT widely accepted in mathematical proofs where quantifiers are usually (always?) restricted by possibly empty sets or predicates. And free variables are introduced only by axioms, premises or existential specification/instantiation. – Dan Christensen Jul 20 '24 at 04:30