How does predicate logic handle contradictory statements about something that does not exist?

Question

Let

• p denote ringing telephone
• s denote someone

• A denote answered

  1. ¬∃x(Px)
  2. ∀x∃y((Px⋀Sy) →Ayx)
  3. ∀x¬∃y((Px⋀Sy) →Ayx)

So,

1. There are no ringing telephones.
2. Every ringing telephone was answered by someone.
3. Every ringing telephone was not answered by someone.

Those three statements could all be true. (I believe 'vacuously true' is the term.) How does propositional logic treat them? How would one prove they are not contradictory?

Answer 1

No, they are not "contradictory", or inconsistent.

Your translations are quite a bit off, which I think only adds to your confusion. (Taking the necessary time to create an appropriate key is essential.)

Key:

Let $Px $ denote "x is a phone".

Let $Sx$ denote "x is a person"

Let $Rx$ denote "x is ringing"

Let $Axy$ denote "x is answered by y".

Using the above key:

$(1)\quad \lnot \exists x(Px \land Rx)$.

$(2)\quad \forall x ((Px \land Rx) \implies \exists y(Sy \land Axy))$.

$(3)\quad \forall x ((Px \land Rx) \implies \lnot\exists y(Sy \land Axy))$.

The statements are not inconsistent, since each of $(2), (3)$ must be true because by $(1)$, the antecedent in each is false.

Answer 2

First, this if first-order logic and not propositional logic (as pointed out in the comments).

There is no contradiction in these three statements being true.

You seem to consider that 3 is the negation of 2, but actually the negation of 2 is

$ \exists p \forall s \neg(A~sp) $