A question on logic - where intuition can fail

Suppose I have 2 predicates $P(x)$ and also $Q(x)$, such that $\overline{P(x)\wedge Q(x)}$ holds for all $x$.

Currently, if $\displaystyle \bigwedge_{x\in A}P(x)$ for a set $A$, it has to be absolutely real, that $\displaystyle \overline{\bigwedge_{x\in A}Q(x)}$ by intuition. Yet intend, that $A = \{\;\;\}$, after that clearly both $\displaystyle \bigwedge_{x\in A}P(x)$ and also $\displaystyle\bigwedge_{x\in A}Q(x)$.

Where is my blunder? I presume it remains in the presumption that $$\bigwedge_x\overline{P(x)\wedge Q(x)} \rightarrow \left(\bigwedge_{x\in A}P(x) \rightarrow \overline{\bigwedge_{x\in A}Q(x)}\right)?$$

¢ *Note: $\displaystyle \overline{P(x)\wedge Q(x)} \equiv \lnot(P(x)\land Q(x))$¢ $\displaystyle \quad\quad\quad\quad \overline{\bigwedge_{x\in A}Q(x)} \equiv \lnot \left(\bigwedge_{x\in A}Q(x)\right)$

2
2022-06-07 14:38:00
Source Share

You are offered $\forall x \lnot(P(x)\land Q(x))$, which coincides as $\forall x \ \lnot P(x) \lor \lnot Q(x)$. After that you are offered $\forall x \in A\ P(x)$. You are appropriate that you can after that acquire $\forall x \in A \lnot Q(x)$. If $A$ is vacant, it is additionally real (with no of the coming before) that $\forall x \in A \ Q(x)$. But also for vacant $A$, this is not an opposition, as it increases to $\forall x (x \in A) \implies Q(x)$ As the antecedent is constantly incorrect, the effects is constantly real.

2
2022-06-07 15:10:35
Source

Your intuition is based upon the set $A$ having some nontrivial dimension. Yet as you keep in mind, the intuition is incorrect if $A$ is vacant.

You just require to transform your intuition a little. If $P$ holds for all $x \in A$, after that we understand that $Q$ can not hold for any kind of $x \in A$. That is, $\bigwedge_{x\in A} \lnot Q(x)$.

The incorrect intuition remains in changing $bkslshbigwedge _ xbkslshin A bkslshlnot Q (x) bkslsh bkslsh bkslsh bkslshrightarrow bkslsh bkslsh bkslsh bkslshlnot bkslshbigwedge _ xbkslshin A Q (x)$. This action is just proper if $A$ is nonempty, yet also after that it is a weak action (for huge $A$ the left hand side is a solid declaration while the right-hand man side is reasonably weak), so you would certainly do ideal to forget this type of makeover bdsh attempt to eliminate it from your instinctive arsenal!

The proper makeover would certainly be that of De Morgan: $bkslshbigwedge _ xbkslshin A bkslshlnot Q (x) bkslsh bkslsh bkslsh bkslshrightarrow bkslsh bkslsh bkslsh bkslshlnot bkslshbigvee _ xbkslshin A Q (x)$.

3
2022-06-07 15:10:32
Source

When I TA had actually a training course which covered comparable subjects we especially needed for the framework to be nonempty.

This remains in order to stay clear of the really trouble that you describe in your inquiry.

Actually, the presumption that you made is really real if you call for the framework to have a nonempty cosmos. Allow us confirm it promptly:

Suppose that for all $x\in A$ we have $\overline{P(x)\land Q(x)}$, which for all $x\in A,\ P(x)$ holds.

Allow $x$ be an approximate component of $A$, after that $\overline{P(x)\land Q(x)}$ which by DeMorgan amounts $\overline{P(x)}\lor\overline{Q(x)}$. We think that $P(x)$ is additionally real, consequently it has to be that $\overline{Q(x)}$.

Given that this $x$ was approximate it holds for all $x\in A$, i.e, $\bigwedge_{x\in A}\overline{Q(x)}$

1
2022-06-07 15:08:37
Source