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I'm reading from Dirk Van Dalen's Logic and Structure and in that the following theorem occurs

Let $A$ be a property, then $A(\varphi)$ hold for all $\varphi \in PROP$ if

  1. $A(p_i)$, for all i and $A(\bot)$
  2. $A(\varphi)$, $A(\psi)$ $\implies$ $A( ( \varphi \square \psi) )$
  3. $A(\varphi) \implies A ( \neg (\varphi) )$

Now, I have few doubts regarding the way things are written:

  1. Why is he writing just "holds" not "holds true"? For example: "then $A(\varphi)$ holds", it should be written "then $A(\varphi)$ hold true"
  2. Why the same symbol $\varphi$ in the actual statement and then in point 2 and 3? I think he should use different symbols.

Now, the proof of it goes like this:

Let $X = \{ \varphi \in PROP | A(\varphi) \}$, then $X$ satisfies (i), (ii), (iii) of definiton 1.1.2. So, $PROP \subseteq X$, i.e. for all $\varphi \in PROP$ $A(\varphi)$ holds.

Can someone please explain the proof to me? I really have no idea, first he assumes $X$ to be the set of those propositions which are in PROP and for whom $A(\varphi)$ is true, well this clearly shows that $X$ will be a subset of $PROP$ but just in the next line he shows that $PROP$ is a subset of $X$. I'm quite confused.

DEFENSE: Other questions do exist for the page and theorem of book but their doubts are different from mine.

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I'm not an English speaker, but I think that in mathematics "it holds" is a usual abbreviation for "it holds true".

Th.1.1.3 (Induction Principle) is a standard expression of Structural Induction.

In Mathematical Induction we say that a statement $P(n)$ holds for every natural number $n = 0, 1, 2,\ldots$.

In the same way, the theorem states the Base case for atomic propositions, and the Inductive clauses corresponding to each connective.

Regarding specifically $\lnot$, the clause reads: for a formula $\varphi$ whatever, if it has property $A$, then also the formula $(\lnot \varphi)$ has property $A$.

The variable $\varphi$ is used in the same way as we use the numerical variable $n$ in Mathematical Induction.


How the proof works?

Mathematical Induction works this way: consider the set $P = \{ n \mid P(n) \text { is true } \}$. By definition: $P \subseteq \mathbb N$.

Now prove that $0 \in P$ and that if $n \in P$ then also $n+1 \in P$. If so, we can apply Induction to conclude that every $n$ belongs to $P$, i.e. that $\mathbb N \subseteq P$.

Cooking them together, we have that $P = \mathbb N$.

In the same way, given a property $A$ of formulas, we consider the set $X = \{ \varphi \in \text {PROP} \mid A(\varphi) \text { is true} \}$. We have: $X \subseteq \text {PROP}$.

Then we use Structural Induction, proving that every atom is in $X$ (clause (i)) and that clauses (ii) and (iii) are satisfied.

Thus, by induction, we conclude that $\text {PROP} \subseteq X$, and finally:

$X=\text {PROP}$.

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  • $\begingroup$ That’s an awesome explanation. When you compared it with $n$ and $\mathbf{N}$ things became quite clear. Thanks. $\endgroup$
    – Delta Psi
    Commented May 24, 2021 at 10:33
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    $\begingroup$ @DeltaPsi - you are welcome :-) $\endgroup$ Commented May 24, 2021 at 10:33

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