How to express statement in $\mathcal{L}_A$?

Question

The statement,

"Zero is less than the successor of any number".

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To my understanding, we express this with in $\mathcal{L} _A$ using a constant ($\textbf{0}$), a function ($\textbf{S}$), addition (+), multiplication (*), and relation symbol (<).

I am confused a bit with the syntax of First-Order Logic.

My attempt:

$<0S0$

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Does this seem correct? I believe I am representing "the successor of any number" incorrectly.

Answer 1

It looks like you're using Polish notation; for what it's worth, this is not the common notation these days.

What you've written says in plain English

$0$ is less than the successor of $0$.

This is not what you want; you want to say

$0$ is less than the successor of any number.

To do this, you're going to need to use a quantifier:

For every $x$, $0$ is less than the successor of $x$.

In symbols (and using your notation), this is $$\forall x(<0Sx).$$

More commonly this would be written as $$\forall x(0<S(x)).$$ Note the extra parentheses ("$S(0)$" instead of "$S0$") and the infix notation for "$<$." One virtue of this is that it flows a bit more naturally when read aloud: "For every $x$, $0$ is less than the successor of $x$" rather than "For every $x$, is less than ($0$, successor $x$)." Obviously this is ultimately subjective, but I think it's worth pointing out the difference.

Answer 2

$$\forall x : 0 < Sx$$

It common for $a < b$ to be an abbreviation for $a \le b \land \lnot (a = b)$

$$\forall x : 0 \le Sx \land 0 \ne Sx$$

because $a \le b$ can then be an abbreviation for $\exists c . a + c = b$ :

$$\forall x . \exists y. 0 + y = Sx \land \lnot (0 = Sx)$$