Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here's the simple grammar for propositional logic I'm using:

  • For all $n \in \mathbb{N}$, $P_n$ is a WFF (Well formed formula).
  • If $\phi$ and $\psi$ are WFF's then $(\phi \rightarrow \psi)$ is a WFF.
  • If $\phi$ is a WFF then $\neg \phi$ is a WFF.
  • Nothing else is a WFF.

My question is this: What is the cardinality of the set$\{\phi: \phi \text{ is a WFF} \}$? My intuition is that it is $\aleph_0$, but I'm having a hard time seeing why.

share|cite|improve this question
Well, write it as a countable union of countable sets... – Zhen Lin Dec 11 '13 at 3:15
I agree with @Zhen. Let $\Phi_0 = \{P_n : n \in \mathbb{N}\}$ and define $\Phi_{i+1} = \Phi_{i} \cup \{\phi \rightarrow \psi : \phi,\psi \in \Phi_i\} \cup \{\neg \phi : \phi \in \Phi_i\}.$ Check that the union of $\Phi_i$ really equals the set of all WFF, and then use basic principles of cardinal arithmetic to deduce that $|\bigcup_{i \in \mathbb{N}} \Phi_i| = \aleph_0$. – goblin Dec 11 '13 at 3:27

HINT: Consider your alphabet $\{P_n\mid n\in\Bbb N\}\cup\{\rightarrow,\lnot\}$. That is a countable set. Every WFF is a finite string over this alphabet. What is the cardinality of all finite strings over a countable alphabet?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.