# Mixing predicate logic with propositional logic

In this paper in legal philosophy (https://doi.org/10.3790/rth.40.1.1) on p. 25, the author writes:

P(x) ↔ (x → ¬h)

This is supposed to be a formalisation of the following statement: "Every utterance of an opinion x is legal as long as it doesn't violate someone's honour." ("Jede Meinungsäußerung x ist rechtmäßig, solange sie nicht gegen das Recht der persönlichen Ehre (h) verstößt.") The author explains that "x=an arbitrary utterance, h=honour"). It is not clear to me what "P" stands for but I assume it is supposed to mean "legal".

It seems to me there are several issues here.

1. The author is mixing predicate logic with propositional logic. "x" is either a variable ranging over utterances ∈ D , or it is a proposition ∈ {True, False}. It seems on the left it is supposed to be the former, on the right the latter. I think therefore, the formula is not well formed. Do I fail to understand it because it is indeed nonsense because it is not a wff? I think, it should rather be something like: P(x) ↔ ¬H(x)

2. However, this formula still does not seem to capture the meaning of the ordinary language sentence appropriately. Rather, it should be: ∀x(¬H(x) → P(x)) or ∀x(¬P(x) → H(x)) respectively. That, however, is false, as a matter of fact, because not violating someone's honour is not a sufficient condition for being an admissible utterance. Rather it is a necessary one. Thus, the correct formula would be, I think: ∀x(P(x) → ¬H(x)) or ∀x(H(x) → ¬P(x)) respectively.

3. The negation indicates that "h" (and my predicate "H") represent not "honour", as the author states but rather "violation of honour". At least if "P" means "legal".

I would be grateful if you could clarify whether I am confused or the author (and reviewers) of this indeed respectable journal.

• You are correct the statement should be made in predicate logic Feb 3, 2023 at 14:42
• P(x) is a predicate; thus x is an individual variable. Thus, x → ¬h is wrong. Feb 3, 2023 at 15:07
• having said that, we can use propositional variables in FOL: a possible example is P(x) ↔ (Q(x) → ¬H) Feb 3, 2023 at 15:58
• We can treat them as zero-place predicates: $H()$ This will behave the same as propositional variables in propositional logic, but is in line with the syntax and semantics of predicate logic. Feb 9, 2023 at 15:04
• That said, what they wrote is clearly wrong, and I wouldn't trust anything of what they have to say about logic in that paper. Feb 9, 2023 at 15:09

Every utterance of an opinion $$x$$ is legal as long as it doesn't violate someone's honour
My translation is $$\forall x{\in}O\;\Big(\,¬\exists y{\in}P\,H(x,y)→P(x)\Big);$$ equivalently, $$\forall x{\in}O\,\exists y{\in}P\;\Big(\,¬H(x,y)→P(x)\Big).$$
• $$H(x,y):$$ uttering opinion $$x$$ violates person $$y$$'s honour
• $$P(x):$$ uttering opinion $$x$$ is legal
• Yep, that's not a wff. BTW, my translation uses the shorthand $∀x{∈}X\:F(x)$ to mean $∀x\:[x{∈}X\to F(x)].$ Feb 3, 2023 at 15:20