Are there paradoxical/ counter-intuitive laws in predicate logic? ( beyond the Drinker Paradox)

Question

Preliminary remarks. (1) The term "paradoxical" is not used in a negative sense here. What is " para-doxical" is literally what disagrees with the general and uninformed " opinion": it could be argued that since Galileo, science has defined itself as literally paradoxical ( what could be more paradoxical than the principle of inertia?). (2) I do not claim that the paradoxical aspect of some logical laws is a deep property. It can be easily shown that these logical laws can be reduced to trivial equivalences. For example, the the alledged " mirabilis consequentia" :

(~A --> A ) --> A

is, in fact , equivalent to

~ ( ( ~A --> A) & ~A ) )

<--> ~ ( ~ ( ~A & ~A) & ~A )

<--> ~ ( ~ ~ A & ~ A)

<--> ~ ( A & ~ A)

<--> (~ A v A )

<--> ( A --> A )

Nevertheless, even if their paradoxical aspect is superficial, there is actually a kind of intellectual pleasure to get first acquainted with such laws, and that is why I ask whether you could make me discover some new ones ( at least new to me) that would belong , this time, not to sentential logic , but to predicate logic.

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In sentential logic, there are some curious or astonishing laws, like consequentia mirabilis: $(\lnot A \to A) \to A$; or verum sequitur ad quodlibet: $A \to (B \to A)$.

I've heard about the "Drinker paradox" in predicate logic:

there is a person such that, if this person is drinking, then everybody is drinking.

Are there other curious/paradoxical ou counter intuitive laws in (monadic/polyadic) predicate logic that would be worth knowing?

More generally, it seems to me more difficult to find a list of laws of (monadic/polyadic) predicate logic than it is for sentential logic. Could you please indicate a classic reference containing such a list?

Answer 1

How about the "paradoxical" theorem: For any set $S$ and proposition $P$, we have:

$$\exists x: [x\in S \to P]$$

OK, this is not exactly what you asked for with its reference to set theory. And it is vaguely reminiscent of the Drinker's Paradox.

HINT: Its proof requires the non-existence of the universal set--not a stretch for most mathematicians, I would think.

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FOLLOW-UP (over a year later): Formal proof using a form of natural deduction and some elementary set theory (constructing subsets) here.

Answer 2

The fact that the material implication does not quite match our intuitions regarding the use of the English 'if ... then ...' (because the material implication is defined as a truth-functional operator, but the English conditional really isn't) leads to various Paradoxes of Material Implication

Your verum sequitur ad quodlibet: $A \to (B \to A)$ is a good example of this: you wouldn;t normally say that if $A$ is true then $B \to A$ is immediately true as well, no matter what $B$ is. But, if you look at the truth-table for the $\to$, that is exactly what is the case for the material implication.

The consequentia mirabilis: $(\lnot A \to A) \to A$ is not an instance of this though, and in fact I don't find that one 'paradoxical at all: If $A$ is true when $\neg A$ is true, then clearly that means (proof by contradiction) that $\neg A$ cannot be true, and hence $A$ is true.

Of all Paradoxes of Material Implication, my favorite one is:

$(P \land Q) \to R \Leftrightarrow (P \to R) \lor (Q \to R)$

That one is really unintuitive! For example, fill in:

$P$: you are male

$Q$: you are unmarried

$R$: you are a bachelor

With these, $(P \land Q) \to R$ would be considered true: yes, being an unmarried male suffices for you to be a bachelor

However, $(P \to R) \lor (Q \to R)$ would be considered false: it is not true that being male is sufficient for you are a bachelor, nor is it true that being unmarried is sufficient for you a bachelor.

If you want this as one statement/law:

$((P \land Q) \to R) \to ((P \to R) \lor (Q \to R))$

is a tautology in propositional logic, but expressed in English you wouldn't think it would be.

By the way, here is a proof of equivalence:

$(P \land Q) \to R \Leftrightarrow$

$\neg (P \land Q) \lor R \Leftrightarrow$

$\neg P \lor \neg Q \lor R \Leftrightarrow$

$\neg P \lor R \lor \neg Q \lor R \Leftrightarrow$

$(P \to R) \lor (Q \to R)$

Answer 3

Some people find it paradoxical that the following two statements are logically equivalent: (1) If you're not part of the solution then you're part of the problem. (2) If you're not part of the problem then you're part of the solution.

In particular, (1) has been used to exhort people to become part of a solution; I have not seen (2) used (for this or any purpose).

This is, of course, another example of what Bram28 already mentioned with other examples: Material implication doesn't always match our intuition.