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How do you do universal generalization on a statement with free variables introduced by both universal and existential specification/instantiation? Consider a formal proof in natural deduction of the form:

  1. Start of proof

$~~~~~~~~~\vdots$

  1. $x$ is introduced (fresh) by universal specification ($\forall$-Elimination)

$~~~~~~~~~\vdots$

  1. $y$ is introduced by existential specification ($\exists$-Elimination)

$~~~~~~~~~\vdots$

  1. $Q(x,y)$ is derived at this point

  2. Can we now infer that $\forall a: \exists b: Q(a,b)$? What is the rule/s?

Assume no free variables other than x and y. And that no assumptions were introduced or discharged after line 10.

EDIT

More concretely, suppose we have...

  1. $\forall a: P(a)~~~$ (Premise)
  2. $\exists b:Q(a)~~~$ (Premise)
  3. $P(x)~~~$ ($\forall$-Elim, 1)
  4. $Q(y)~~~$ ($\exists$-Elim, 2)

Can I now use natural deduction (FOL) to obtain: $\forall a: \exists b: (P(a) \land Q(b))$? Are there any "side conditions," "dependencies," etc. introduced by $\exists$-Elim here that must be taken into consideration?

Dan Christensen
  • 14,529
  • There is no "Existential Specification" in ND... – Mauro ALLEGRANZA Oct 06 '20 at 13:54
  • @MauroALLEGRANZA Same as $\exists$-Elimination. – Dan Christensen Oct 06 '20 at 16:39
  • Can you give a more concrete example of which formulas you come from and get to by your steps 10 and 20? – Natalie Clarius Oct 06 '20 at 22:19
  • @lemontree See edit. – Dan Christensen Oct 07 '20 at 14:18
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    Step 5. $P(x) \land Q(y)$ [by $\land$-intro]. 6. $\exists b [P(x) \land Q(b)]$ [by $\exists$-intro: no side conditions]. Now the formula has no occurrence of parameter $y$ used in step 4 for $\exists$-Elim. Thus, we can "close" the sub-proof, discharging the assumption made in step 4. because we have correctly derived $\exists b [P(x) \land Q(b)]$ by $\exists$-E. 7. $\forall a \exists b [P(a) \land Q(b)]$ by $\forall$-I. There are no undischarged assumptions with $x$ free: thus, the side condition for $\forall$-I is satisfied. – Mauro ALLEGRANZA Oct 08 '20 at 09:30
  • @MauroALLEGRANZA AFAICT "side conditions" never show up in mathematical proofs. Why is that? Where can I find a straightforward explanation of them? – Dan Christensen Oct 08 '20 at 14:46
  • Side conditions (aka: proviso) are necessary for formal derivation, because (in principle) they can be performed by a machine. In human proofs, we have intuition: I imagine that every "trained" mathematicians can easily avoid the fallacy: "Socrates is a Philosopher; therefore everything is a Philosopher". – Mauro ALLEGRANZA Oct 08 '20 at 14:54
  • @MauroALLEGRANZA Somehow I have managed to avoid anything that looks like a "side condition" in my formal system of logic. Unless I have subtly built them into my software. To use your example I cannot infer from $P(s)$ that $\forall x (P(x))$. (I do not allow the introduction of new free variables by Universal Instantiation ($\forall$-Elim).) – Dan Christensen Oct 08 '20 at 15:42

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