Universal Quantifier Representation (Conditional Statement): Implication, Semantic Consequence, Syntactic Consequence

Question

I have seen $\forall x [P(x)\rightarrow Q(x)]$ written in multiple different forms from different authors. Which of the following are correct if any (as listed below)?

$\forall x [P(x)\rightarrow Q(x)]\equiv P(x)\Rightarrow Q(x)$
$\forall x [P(x)\rightarrow Q(x)]\equiv$ P(x), implies Q(x)
$\forall x [P(x)\rightarrow Q(x)]\equiv P(x) \models Q(x)$
$\forall x [P(x)\rightarrow Q(x)]\equiv P(x) \vdash Q(x)$

No one. $\forall x \ [P(x) \to Q(x)]$ reads: "every $P$ is $Q$". See Categorical proposition. — Mauro ALLEGRANZA, Jun 28 '17 at 12:52
The "syntactic consequence" relation: $P(x) \vdash Q(x)$ reads: "from (formula) $P(x)$, (the formula) $Q(x)$ is derivable". See Proof calculus (or Proof system). — Mauro ALLEGRANZA, Jun 28 '17 at 12:53
The "semantic consequence" relation: $P(x) \vDash Q(x)$ reads: "(formula) $P(x)$ logically implies (the formula) $Q(x)$". See Logical Consequence. — Mauro ALLEGRANZA, Jun 28 '17 at 12:54
@MauroALLEGRANZA: I was hoping you were going to answer my question when I posted this. I appreciate your help! — W. G., Jun 28 '17 at 12:57
You can see the post deduction theorem intuition for relation between $\vdash$ and $\to$. — Mauro ALLEGRANZA, Jun 28 '17 at 13:14
@MauroALLEGRANZA: I know this question is unrelated, but could you rewrite the deduction theorem as this $(\Delta \cap { A } \vdash B) \models (\Delta \vdash A\rightarrow B)$ then? — W. G., Jun 28 '17 at 13:36
No; it must be $\Delta \cup { A }$, because the set of premises is made of the formulas in $\Delta$ "plus" the formula $A$. $\Delta \cap { A }$ can be at most $A$ (or maybe $\emptyset$). — Mauro ALLEGRANZA, Jun 28 '17 at 13:49
The Ded Th reads: "if ..., then ..." at the meta-level: if we have a derivation (in the proof system) of $B$ from the set of premises $\Delta, A$, we have also (in the sense that we can produce) a derivation of $A \to B$ from $\Delta$. — Mauro ALLEGRANZA, Jun 28 '17 at 13:51

score 3 · Accepted Answer · answered Jun 28 '17 at 13:07

No one.

$∀x [P(x)→Q(x)]$ reads: "every $P$ is $Q$". See Categorical proposition.

The "syntactic consequence" relation: $P(x)⊢Q(x)$ reads: "from (formula) $P(x)$, (the formula) $Q(x)$ is derivable". See Proof calculus (or Proof system).

The "semantic consequence" relation: $P(x)⊨Q(x)$ reads: "(formula) $P(x)$ logically implies (the formula) $Q(x)$". See Logical Consequence.

$⇒$ is ambiguous; sometimes it is used for (logically) implies (i.e. semantic consequence), sometimes for the connective "if..., then..." (i.e. the conditional: →).

score 1 · Answer 2 · answered Jun 28 '17 at 13:48

As Mauro says: no one! One reason for this that applies to all of them is that $\forall x (P(x) \rightarrow Q(x))$ is a sentence, as it is a formula without any free variables, but in all of the 4 cases you list you are dealing with free variables, i.e. formulas, but not sentences.

Of course, to really see that, you need to know what those different symbols mean, and Mauro explained all that, although one immediate clue should have been that in 1 through 4 there are no $\forall x$ being used.

So, maybe a more interesting question would have been: is the formula $P(x) \rightarrow Q(x)$ equivalent to any of 1 through 4? In which case 3 and 4 are still definitely not the same (see Mauro's answer), while for 1 and 2 they could be the same; by 'implies' we sometimes mean 'logically implies' (which would be like 3), but other times we say 'P implies Q' to mean 'if P then Q', in which case it would be the same as $P \rightarrow Q$. And, as Mauro says, the same is true for $\Rightarrow$.

Thank you as well. Your help is always appreciated! – W. G. Jun 28 '17 at 14:37 — W. G., Jun 28 '17 at 14:37

Universal Quantifier Representation (Conditional Statement): Implication, Semantic Consequence, Syntactic Consequence

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