2

I am currently reading a book on natural deduction, and it states that for universal generalization or $\forall$-introduction, defined as:

$$\frac {\phi[t/x]} {\forall x \phi}$$

The following restrictions apply:

  1. The term $t$ cannot occur in any undischarged assumption of the derivation of $\phi[t/x]$
  2. The term $t$ cannot occur in $\phi$

The first of these restrictions is easy to understand, but for the second I just can't quite understand why it is needed? In the process of going from the top formula to the bottom, you replace all instances of t with x, so therefore $\phi$ should have no $t$'s in it. Therefore, what's the use of a restriction on a situation that can never occur? Can someone give me a (realistic) example of where this restriction may be needed?

esotechnica
  • 777

3 Answers3

2

Consider:

$$\begin{array} {r|ll} % (1) & Pt \iff Pt & \text{Provable} \\ % (2) & \forall x ~ Px \iff Pt & \forall \text{ Intro of (1)} \\ % (3) & \forall y \forall x ~ Px \iff Py & \forall \text{ Intro of (2)} \\ % \end{array}$$

This is clearly an unsound derivation, and the $\forall \text{ intro}$ into step (2) violates restriction 2. Your book is writing UI is a strange way, normally you'd write:

$$\frac{\phi}{\forall x~\phi[x/t]}$$

with only restriction 1. Restriction 2 is trying to get across "you can't replace only some of the (free) $t$, you have to replace all of them". Informally UI tries to capture the concept "if a variable (in this case $t$) has no assumptions made about it, then it could be anything". But there is always the implicit assumption that a variable is itself, which is broken if you only replace some of them.

DanielV
  • 23,556
  • Thanks, I thought that the notation $\phi[t/x]$ meant that you have to replace all t, I will have to check the book and see if this is mentioned or not. – esotechnica Oct 15 '16 at 22:25
  • 2
    @esotechnica The notation $\phi[t/x]$ means that you have to replace all of the $x$. I think of it as "t is over the x, it pushes them down and out". Not that $\phi[t/x]$ is in the assumption, which makes it a bit weird. You aren't doing a replacement on the assumption to get a conclusion, you are doing a replacement on the conclusion $\phi$ to get the assumption. – DanielV Oct 15 '16 at 22:29
  • Yes, just confirmed the book states that replacement must be done everywhere, so restriction 2 seems unnecessary. And yes, I agree with you that the (usual) form you have used is much clearer. – esotechnica Oct 15 '16 at 22:35
  • @esotechnica No, restriction 2 is necessary. I think you are confusing both the variable which is being replaced, and the fact that the deduction goes from assumption to conclusion but the replacement (as written in your book) goes from conclusion to assumption. – DanielV Oct 15 '16 at 22:40
  • OK, I think I get what you mean, but it seems like a confusing way to explain it (the textbook I mean). I think I will take your definition of $\forall$-intro with restriction 1 only, it hurts my brain less! – esotechnica Oct 15 '16 at 22:49
0

For the restriction 1., consider as $\phi$ the formula : $(x=0)$ :

1) $t=0$ --- assumption : is $\phi[t/x]$

2) $\forall x \ (x =0)$ --- from 1) by $\forall$I : wrong : $t$ is free in the one-line derivation of $\phi[t/x]$ , i.e. in $t=0$

3) $t=0 \to \forall x \ (x =0)$ --- from 1) and 2) by $\to$I

4) $\forall x \ [x=0 \to \forall x \ (x =0)]$ --- from 3) by $\forall$I : now there is no undischarged assumption; thus the rule is correctly applied

5) $0=0 \to \forall x \ (x =0)$ --- from 4) by $\forall$E.

The same apply for 2.

If we have a derivation $\mathcal D$ ending with e.g. $\phi(t)$, now this formula is not an assumption, but a similar counter-example shows that the proviso $t$ not occurring in $\phi$ is needed.

A quite "unnatural" example can be : $\phi$ is $(x \ne 0) \lor (x \ne 1)$.

In first-order arithmetic, we can prove, by $\lor$-intro : $(0 \ne 0) \lor (0 \ne 1)$.

Thus the following derivation as no undischarged assumptions.

1) $(0 \ne 0) \lor (0 \ne 1)$ --- considering the constant $0$ as $t$

2) $\forall x \ [(x \ne x) \lor (x \ne 1)]$

that is false.

The restrictions "formalize" the simple fact that the constant or variable $t$ must be "new".

See :

Mauro ALLEGRANZA
  • 94,169
  • I don't understand how restriction 2 can be needed? No formula $\phi$ can ever contain a term t, because the process of adding the quantifier to get $\forall x \phi$ has just removed all instances of t from $\phi$ and replaced them with x! Isn't this how it works? I really need a concrete example to demonstrate this. I am completely confused. – esotechnica Oct 15 '16 at 09:03
0

If the variable $t$ is allowed to occur in $\phi$, you allow $\phi\equiv \forall x\exists t(x\ne t)$. Now what is $\phi[t/x]$?

Hagen von Eitzen
  • 374,180
  • $\phi[t/x]$ would be $\forall x \exists t(t \neq t)$ Which I can see is nonsense but I am struggling to put this into a context that relates to the universal generalization rule. Can you construct an argument that somehow produces a false deduction when this second condition is breached? – esotechnica Oct 15 '16 at 08:07
  • $\phi[t/x] = \forall x \exists t (x \ne t)$. Replacement only refers to free variables. I believe this is an incorrect counterexample. – DanielV Oct 15 '16 at 21:38