# Is falsity both true and false? [closed]

I have been thinking about the liar's paradox a lot lately (which I will post about soon), and I've run into a seemingly unrelated paradox. Consider this sentence: Falsity is both true and false. Symbolizing the sentence one way yields: F=T&F. This is a true statement because T&F simplifies to F. Symbolizing another way yields: (F=T) & (F=F). This statement is false. So my question is whether the sentence is a legitimate paradox or whether the different truth values come from an ambiguity in the English language. Thanks for helping!

• Liars Paradox is game theory, not philosophy. Oct 15, 2021 at 5:20
• @SwamiVishwananda That is definitely not the received view within mainstream philosophy.
– E...
Oct 15, 2021 at 5:42
• This is not a paradox. What you discovered is that A ≡ B&C and (A ≡ B)&(A ≡ C) are not logically equivalent propositions, where "≡" is logical equivalence. Specifically, when A is F, B is T and C is F we get different truth values. So (A ≡ B&C) ≢ (A ≡ B)&(A ≡ C), equivalence does not distribute over conjunction. This should not be surprising. The second one implies that B ≡ C, but there is no such requirement in the first. So you are "symbolizing" two different things and getting a mismatch. Perhaps they do sound similar as said in English, so one could attribute it to language ambiguity. Oct 15, 2021 at 9:17
• F=T&F is not an English sentence. It is a formula in Boolean algebra and the "rules of the game" are exactly those: False & whatever is False. This does not mean that False is also "whatever". Oct 15, 2021 at 16:15

It is true that ⊥ is equivalent to ⊤∧⊥, so that ⊥↔(⊤&⊥) is true. However, we cannot distribute the biconditional across the conjunction; 𝐴↔(𝐵&𝐶) is not in general equivalent to (𝐴↔𝐵)&(𝐴↔𝐶), and in particular we cannot conclude from the above that the sentence (⊥↔⊤)&(⊥↔⊥) is true. Indeed, this latter sentence is false.

In terms of the natural language you've employed, the issue is that "Falsity is both true and false" can be interpreted in two ways:

• as "the conjunction of 'true' and 'false' is false," which is true; or

• as "falsity is true and falsity is false," which is false.

These two interpretations are not the same, and the conflation of them amounts to the claimed distributivity above.

• I interpreted the OP’s assertion not as distributing the biconditional, but going from T & F to (F <-> F) & (F <-> T), by substituting (F <-> F) for T and (F <-> T) for F. The OP did note your “latter sentence” is false (assuming he meant the biconditional rather than equality). Oct 15, 2021 at 5:09
• @JustSomeOldMan I've added a bit of explanation about where I'm getting this interpretation of the OP. (Of course ultimately this is pretty good evidence that the OP should explain themself more clearly.) Oct 15, 2021 at 5:12
• I see, and I agree. Oct 15, 2021 at 5:16
• Out of curiosity, why the downvote? Oct 15, 2021 at 5:38
• It wasn’t me. I didn’t downvote your answer nor the original question. Oct 15, 2021 at 6:02

Your proposition is undefined until you specify what logic it is under. It seems, at a glance, you want to work in propositional logic, but propositional logic does not have equality. Most logical systems don’t. The closest thing in propositional logic is the biconditional “<->”.

Now, F <-> (T & F) is True, but this does not mean that F and (T & F) are the same. It merely means they have the same truth value.

“Falsity” is not both True and False. The truth value of F has the same truth value as (T & F) after simplification, but that is a much weaker assertion.