How can I formalize the following sentence into first-order logic?

Question

Every by $10$ divisible number is ending by the digit $0$.

My solution:

let be $N(x)$ a natural number, which is divisible by $10$,

let be $L(x)$ the last digit of a number

$( \forall x)( N(x) \implies L(x)) $

and the negation of it is:

$ (\exists x)( \neg N(x) \implies \neg L(x)).$

Answer 1

Your first symbolization is correct, but for the negation of it, please note that the $\neg$ does not distribute over the $\Rightarrow$. That is:

$$\neg (P \Rightarrow Q) \not \equiv \neg P \Rightarrow \neg Q$$

But rather:

$$\neg (P \Rightarrow Q) \equiv \neg (\neg P \lor Q) \equiv \neg \neg P \land \neg Q \equiv P \land \neg Q$$

As such:

$$\neg ( \forall x)( N(x) \implies L(x)) \equiv ( \exists x)\neg ( N(x) \implies L(x))\equiv ( \exists x)( N(x) \land \neg L(x))$$

Answer 2

For your latter case, the negation would be:

$\exists x$ $(N(x) \wedge \neg L(x) )$

Answer 3

Always remember:

• $\forall x~(P(x)\to Q(x))$ says: "Any thing, if $P$, will be $Q$".
• $\exists x~(P(x)\wedge Q(x))$ says: "Some thing, is $P$, and also $Q$".

Also recall that the negation of implication is a conjunction vis: $\neg (A\to B)\iff A\wedge\neg B$

So:

• "Any (natural)number, divisible by 10, will end in the digit zero."

  $$\forall x~(N(x)\to L(x))$$

• "Not every (natural)number, divisible by 10, will end in the digit zero."

  "Some (natural)number, divisible by 10, does not end in the digit zero."

  $$\exists x~(N(x)\wedge\neg L(x))$$