Why if for all x, p(x) and for all x not p(x) is not a contradiction?

Question

$\forall x: p(x) \equiv1$ and $\forall x: \neg p(x) \equiv 1$ is not a contradiction?

I have this doubt after a logic contest, and I cant see why. My thoughs was that this is not a contradiction because maybe the TRUE set of these two proposition may be the empty set, but Im not sure if this is the right answer.

I will appreciatte any coment.

Answer 1

I think that there is a "terminological" issue here ...

In "modern practice", we say that :

a formula $\phi$ is a contradiction, or inconsistent, if it is unsatisfiable.

This definition can be easily applied to a set of formulae.

In this sense, the two formulae $\forall x p(x)$ and $\forall x \lnot p(x)$ form an unsatisfiable set of formulae (in the "usual" semantics that does not admit empty universes).

But there is an "older" meaning of contradictory : see The Traditional Square of Opposition :

Two propositions are contradictory iff they cannot both be true and they cannot both be false.

Two propositions are contraries iff they cannot both be true but can both be false.

According to these definitions, the two sentences are contraries, because they can be both false.

Answer 2

Well I suppose, in First-order logic one could consider an empty universe (or domain of discourse). In this case, any statement over an $\forall$ quantifier is true. So you could say that there exists a universe such that both statements are true and thus they are no contradictions.

This, however, relies very strongly on which definition one uses since mostly universes are assumed to be non-empty...

Answer 3

The answer may depend on the universe of discourse. If $x$ is a variable for unicorns, then "All unicorns have blue eyes" and "all unicorns have non-blue eyes" are both true. However, we paractically always assume a nonempty universe of discourse (and some rules of natural deduction rely on that). In that case, from $\forall x p(x)\equiv 1$ we infer $p(a)\equiv 1$, from $\forall x \neg p(x)\equiv 1$ we infer $\neg p(a)\equiv 1$, and these two are in contradiction.