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.

  • $\begingroup$ You are correct the statement should be made in predicate logic $\endgroup$ Feb 3 at 14:42
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    $\begingroup$ P(x) is a predicate; thus x is an individual variable. Thus, x → ¬h is wrong. $\endgroup$ Feb 3 at 15:07
  • $\begingroup$ having said that, we can use propositional variables in FOL: a possible example is P(x) ↔ (Q(x) → ¬H) $\endgroup$ Feb 3 at 15:58
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    $\begingroup$ 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. $\endgroup$ Feb 9 at 15:04
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    $\begingroup$ 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. $\endgroup$ Feb 9 at 15:09

1 Answer 1


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
  • $\begingroup$ Thank you, both of you. Can you confirm that P(x) ↔ (x → ¬h) is not a well-formed formula? $\endgroup$ Feb 3 at 15:10
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    $\begingroup$ Yep, that's not a wff. BTW, my translation uses the shorthand $∀x{∈}X\:F(x)$ to mean $∀x\:[x{∈}X\to F(x)].$ $\endgroup$
    – ryang
    Feb 3 at 15:20
  • $\begingroup$ Welcome to this site! Accepting ✔ and upvoting ▲ answers are good ways to say thanks; these actions signal resolution, prevent the page from being bumped, score points, and influence the site's search results, cleanup activities, and other behind-the-scenes processes. $\endgroup$
    – ryang
    Feb 28 at 5:56

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