How does one show that the set of all axiomatic proofs (in propositional logic) is countably infinite?

Question

I'm supposing we have to show that there an injective function from the set of all axiomatic proofs to the set of all natural numbers? If so, what would such a function look like?

Any help would be greatly appreciated!

Answer 1

It's easier to note there are countably many strings of finite length in any language with a countable alphabet. Indeed,$$\kappa\le\aleph_0\to\sum_{n\ge0}\kappa^n\le\sum_{n\ge0}\aleph_0\le\aleph_0^2=\aleph_0.$$

Answer 2

Let $\Sigma$ be a set of symbols that contains all of the (countably many) propositional variables, all of the (probably finitely many) logical operators, and perhaps some brackets and a comma for good measure.

Then every propositional formula—and, in fact, every list of propositional formulae—is written using symbols that are elements of $\Sigma$.

Evidently $\Sigma$ is countable. Therefore $\Sigma^2 = \{ \sigma \tau : \sigma,\tau \in \Sigma \}$ is countable, since it is in bijection with $\mathbb{N} \times \mathbb{N}$, which is countable. A straightforward induction reveals that $\Sigma^n$, the set of strings from $\Sigma$ of length $n$, is countable for all $n \in \mathbb{N}$.

But then $\Sigma^* = \bigcup_{n \in \mathbb{N}} \Sigma^n$, the set of all finite strings from $\Sigma$, is a union of countably many countable sets, so is countable.

An axiomatic proof can be viewed as a (finite) sequence of propositional formulae, each following from the last by some axiom or inference rule. But then an axiomatic proof is simply an element of $\Sigma^*$, built by separating each formula in the sequence by a comma.

So the set of all axiomatic proofs is a subset of $\Sigma^*$. Since $\Sigma^*$ is countable, the set of all axiomatic proofs is countable.

Answer 3

The other answers described why the set of proofs must be countable.

If you are curious what the function would look like: you can describe a well-order on axiomatic proofs as strings of some language by $L<M$ if $\#L<\#M$ or if $\#L=\#M$ and $L$ occurs before $M$ in lexicographic order (based on some order of the finite alphabet of your letters and symbols that make up your proof).

Once you have that, then $f(0)$ is the least member of $\Sigma$, $f(1)$ is the least member of $\Sigma\setminus \{f(0)\}$, and so on.

Answer 4

$A_n:=$

{ $a_n| a_n$ is a string of length $n$, means : there are $n$ positions each of which filled with an element of a finite extended alphabet, say letters, numbers, symbols, etc.}

$A_n$ is finite.

$\bigcup_{n \in \mathbb Z^+} A_n$ , a countable union of finite sets is countable.

$B:= ${$b|$ $b$ is an axiomatic proof}.

Then

$B \subset \bigcup_{n \in \mathbb Z^+}A_n $

$B$ as a subset of a countable set is countable.