# Predicate logic and always?

How do I translate a sentence like this into predicate logic?

Always if an amateur chef bakes a burnt cookie, then nobody eats that burnt cookie.

My attempt is something like this ∀[chef’(x)∧ amateur'(x) ∧ burnt'(y) ∧ cookie'(y) ∧ bakes'(x, y)]→¬∃(z)[person'(z) ∧ eat'(z, y)]]

I am wondering how to do the adjectives and quantify over case with always while still using "nobody." Help appreciated:)

EDIT: This is considering Lewis' theory of adverbs of quantification, which uses unselective quantifiers to quantify over case. Here is the article: users.ox.ac.uk/~sfop0776/LewisQA.pdf I am sorry I didn't initially post it!

• I suggest starting with something simpler like "All men are mortal". Dec 18, 2019 at 17:46
• ∀x[men'(x)→mortal'(x)] I understand that. I am trying to understand quantifying over case. Thank you:) Dec 18, 2019 at 17:52
• So your formula would be equivalent to ∀x∀y[...], right? I've seen people reify time along the lines ∀z[Time(z)]. Dec 18, 2019 at 18:07
• In a way yes. I am trying to understand how Lewis would want to translate this into predicate logic with his ideas in Adverbs of Quantification. He proposed that unselective quantifiers can capture this meaning, so ∀formula would be true iff F is true under every admissible assignment to all variables free in F. I am confused with the word nobody, should I bind it, or leave it free? Dec 18, 2019 at 18:10
• Your predicate logic translation looks good to me. I'm not clear what problem you see with it. Dec 18, 2019 at 19:54

Predicate logic is an approximation of some functions of language, and does not cover all use cases. Traditionally, it doesn't really account for time (with just entities, there isn't a good way to distinguish past tense from future tense). I've seen some analyses that use predicate logic as a base add in a "time" variable to every verb (BAKE(x, y, t) = x bakes y at time t). In this case, that always can be represented something along the lines of λP.(∀t.P(t)) (always is a function where you put in a predicate, and it then says "for all times, that predicate is true")

• That makes more sense! Thank you! But let's say I still want to understand what Lewis is saying because his article confused me. Would nobody be bound, or would it be free because always is over all cases? Dec 18, 2019 at 18:15
• I don't know which Lewis or which article you are referring to, but "nobody" is free here because it does not reference any variables bound in the scope it is in (it doesn't require the x or y which are bound at that scope) Dec 18, 2019 at 18:52

Your formula is not well-formed: There needs to be a variable after the ∀. Specifically, it should be those variables that you want to universally quantify over. Your variables `x` and `y` are free (= not bound by any quantifier) but shouldn't be.

Fixing this is the answer to your problem: "Always" in this context is best translated not as something temporal, but as "For any amateur chef and any burnt cookie, if the amatuer chef boke that cookie, then ..."

So all you need to do is universally quantify over `x` and `y`, and extending the scope of the universal quantifiers over the whole formula instead of only up until the implication (since the `y` in `eat(x,y)` should belong to the other bound `y`s):

``````∀x∀y[chef’(x) ∧ amateur'(x) ∧ burnt'(y) ∧ cookie'(y) ∧ bakes'(x,y) →
¬∃(z)[person'(z) ∧ eat'(z,y)]]]
``````
• While normally I would agree with you, and I appreciate your thoughtful response, but Lewis in "Adverbs of Quantification" argues for unselective quantifiers in sentences like always.users.ox.ac.uk/~sfop0776/LewisQA.pdf I am trying to understand how this would work with an unselective quantifier. Dec 18, 2019 at 19:43
• This is an important enough point that your question suffers from not making this explicit in the question. Otherwise, your formulation looks a bit clueless w.r.t. elementary logic, and we don't know what you know but aren't saying. Dec 18, 2019 at 20:48
• I am sorry I am stupid. Dec 18, 2019 at 22:07