Application of Conditionals to Sets

Question

Our biggest issue is how to make sense of conditionals in the Truth Table Format, and applying them to understand sets. We tried coming up with an easy example in excel for our own purpose.

Let $P:= x\gt 2$, and $Q: = x^2 > 4$.

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The red highlighted figures are incorrect because the converse and inverse are supposed to be equivalent, and contrapositives of conditionals are supposed to be equivalent to the conditional. However, we can't seem to make sense of this.

This is for a study group. Let us know!

Thanks!

Answer 1

You can't fully represent the information in propositional logic (aka, sentential logic).

What you need to fully capture the information is to use predicate logic with quantifiers:

E.g. Let $P(x)$ denote $x\geq 2$. Then $\lnot P(x)$ will be, essentially, $x\lt 2$.

Let $Q(x)$ denote $x^2 \geq 4$, so that $\lnot Q(x)$ becomes $x^2\lt 4$.

Let's make the universe of discourse the set of the non-negative real numbers, $\mathbb R_{x\geq 0}$.

Then you might say something like:

$$\forall x\Big((P(x)\rightarrow Q(x))\land ((\lnot P(x)) \rightarrow (\lnot Q(x))\Big)\tag 1$$

This can be adapted for all reals, but you'd need $P(x): (x \leq -2)$ or $(x\geq 2)$, so $\lnot P(x): -2 < x < 2$.

Likewise you can extend $Q(x)$ over the domain of real numbers: Let $Q(x): x^2 \geq 4$, while $\lnot (Q(x)): 0\leq x^2\lt4$

Then $(1)$ holds as well, given the modified domain and predicates.

Answer 2

Recall that $$P\to Q$$ is always true, unless $P$ is true, and $Q$ is false. Thus $$\lnot Q\to \lnot P$$ is always true, unless $\lnot Q$ is true, and $\lnot P$ is false. But this is indeed equivalent to the above.

Of course none of these is equivalent to $$Q\to P,$$ as this is always true, unless $Q$ is true, and $P$ is false.