So im trying to learn some basic logic and I got stuck on how we use implications with the universal quantifier.

My problem is the following:

$\forall x {\in}\varnothing\; P(x)$ is the same as saying

$\forall x\; \big(x\in\varnothing \to P(x)\big)$

However if I've been given a statement of the form $\forall x\in A \; \big(x\in\varnothing \to P(x)\big)$ should I read this as $\forall x\; \big(x\in A \to x \in \varnothing \to P(x)\big)$

or instead $\forall x\; \big(x\in A \wedge x \in \varnothing \to P(x)\big)$.

Thanks in advance .

  • 4
    $\begingroup$ $(P \land Q) \to R$ is equiv to $P \to (Q \to R)$ $\endgroup$ Mar 22, 2021 at 19:35
  • $\begingroup$ @mauro ALLEGRANZA oh yes now I see the answer was right in front of me thankyou for pointing it out. $\endgroup$ Mar 22, 2021 at 19:43

1 Answer 1


A statement of the form $$\forall x\in A[Q(x)]$$ always means $$\forall x[x\in A\rightarrow Q(x)],$$ by definition (the first statement is just an abbreviation for the second). So, $$\forall x\in A[x\in\emptyset \rightarrow P(x)]$$ means $$\forall x[x\in A\rightarrow (x\in\emptyset\rightarrow P(x))].$$ This is equivalent to $$\forall x[(x\in A\wedge x\in\emptyset)\rightarrow P(x)]$$ though since $x\in A\rightarrow (x\in\emptyset\rightarrow P(x))$ and $(x\in A\wedge x\in\emptyset)\rightarrow P(x)$ are equivalent. I'm not sure if this is what you meant with your second statement since you left out the parentheses.

  • $\begingroup$ are the parentheses required what do the extra parentheses add to the equation $\endgroup$ Mar 22, 2021 at 22:33

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