Understanding $PROP$ set in the book Logic and Structure (Van Dalen).

Question

Working on the book: Dirk van Dalen. "Logic and Structure (Universitext)" (p. 18)

Definition 1.1.2 The set PROP of propositions is the smallest set X with the properties

$ \begin{array}{rl} \rm(i)&p_i\in X(i\in N),\bot\in X,\\ \rm(ii)&\varphi,\psi\in X\Rightarrow(\varphi\wedge\psi),(\varphi\vee\psi),(\varphi\to\psi),(\varphi\leftrightarrow\psi)\in X,\\ \rm(iii)&\varphi\in X\Rightarrow(\neg\varphi)\in X.\\ \end{array} $

I would like to know:

$p_i\in X(i\in N),\bot\in X$

• How can I instantiate this statement when verifiyng a string of symbols belongs to PROP ?
• Is the comma an and connective ?
• What is $N$?
• Why is bottom symbol there ?

$((p \land q) \to p)$

• How can I show this statement belongs to PROP ?

P.S.: I am already aware of similar questions but they do not address my questions, I think.

Answer 1

$p_i\in X(i\in N),\bot\in X$

• How can I instantiate this statement when verifiyng a string of symbols belongs to PROP ?

You can instantiate any propositional symbol or $\bot$: $p_1$ is $\in PROP$ according to this clause, $p_{354}$ is $\in PROP$ according to this clause, $\bot$ is $\in PROP$ according to this clause.

• Is the comma an and connective ?

Yes.

• What is $N$?

It should be $\mathbb{N}$. The $(i \in N)$ just indicates that the $i$'s are running indices to number the propositional symbols.

• Why is bottom symbol there ?

Because it is an atomic formula. Unlike the other connectives, it does not take any other formulas to form a new formula, and thus belongs in the base case together with the propositional symbols.

$((p \land q) \to p)$

• How can I show this statement belongs to PROP ?

Strictly speaking you can't, because according to the definition introduced up to this point, there exist only indexed propositional symbols starting iwth $p$. But it is customary to use $p, q, r$ in practice; if we incoporate these into clause (i), one can prove by induction on the structure of the formula:

1. $p \in PROP$ (by (i), with $p_i = p$).
2. $q \in PROP$ (by (i), with $p_i = q$).
3. Since $p \in PROP$ and $q$ in $PROP$, $(p \land q) \in PROP$ (by (ii), with $\phi = p$ and $\psi = q$).
4. Since $(p \land q) \in PROP$ and $p \in PROP$, $((p \land q) \to p) \in PROP$ (by (ii), with $\phi$ = $(p \land q)$ and $\psi = p$).