# FOL translations (double predicates)

I'm learning first order logic right now and I'm stuck on some translations.

Key:

G(x) : x is a galaxy

M(x) : x is a moon

P(x) : x is a planet

S(x) : x is a star

B(x,y) : x is bigger than y

O(x,y) : x orbits around y

Q1: Everything with a moon orbiting around it is bigger than a planet.

My attempt: Ax (Ey (M(y) /\ O(y,x)) -> Ey (P(y) /\ B(x,y)) )

Q2: Some galaxy is not bigger than all of the moons.

My attempt: Ex (G(x) /\ Ay ~(M(y) -> B(x,y)))

Q3: Any star with a planet orbiting around it is not a galaxy.

My attempt: AxAy((S(x) /\ P(y) /\ O(y,x) )-> ~G(x) )

If you could tell me where I've gone wrong in these questions that'd be appreciated.

• I’m voting to close this question because this is not a forum for logic class homework questions Nov 9 '20 at 6:55
• Welcome to SE Philosophy! Thanks for your contribution. Please take a quick moment to take the tour or find help. You can perform searches here or seek additional clarification at the meta site. Don't forget, when someone has answered your question, you can click on the arrow to reward the contributor and the checkmark to select what you feel is the best answer.
– J D
Nov 9 '20 at 19:24
• This is an excellent example of how to ask a question about doing logic.
– J D
Nov 9 '20 at 19:25

Hi these would be my answers:

Q1: Everything with a moon orbiting around it is bigger than a planet.

(∀x)(∃y)((My & Oyx) ⊃ (∃z)(Pz & Bxz))

I think you would require a third z to indicate the predicate "planet".

Q2: Some galaxy is not bigger than all of the moons.

(∃x)(Gx & (∀y)(My ⊃ ~Bxy))

This would my answer for Q2 although I think your attempt isn't incorrect too.

Q3: Any star with a planet orbiting around it is not a galaxy.

(∀x)(∀y)((Sx & Py & Oyx) ⊃ ~Gx)

Same as yours for Q3.

I'm going to offer an answer to the first of your three to show a strategy for getting from the natural language to an artificial formalism.

Q1: Everything with a moon orbiting around it is bigger than a planet.

I'm going to take 'a planet' as all planets because I'm interested in presumptions in natural language convention. If someone says 'Everything that is a planet is bigger than a mouse', one can infer that one is not talking about some mice, but rather all mice. This is from my reading on principles of implicature, specifically a maxim of cooperation. Strictly speaking, from a formal perspective, the indefinite 'a' should be used more broadly. As I am interested in how natural language translates into symbolic formalisms, I'm taking a slightly different tack to see how things turn out. But breaking problems into pieces is the lesson here.

Right away, one can see the complexity, and one of the best ways of solving complex problems is to reduce them to simpler parts. In this case, we have two predications occurring: is orbiting and is bigger, so let's split that apart:

First Part: O(x,y)

P1. There exists something with a moon orbiting it, or even simpler:
P1'. Some moon orbits something.

That's easy, right? Now, from less formal to more formal.

S1. O(Mx,y), or even more formally...
S1'. ∃x,y : Mx & O(x,y), or even...
S1''. ∃xy OMxy, which is almost incomprehensible.

Note that now, x is bound to the set M and is in a binary relation with y which is unbound.

Second Part: B(x,y)

P2. There exists something bigger than some planet, or more simply...
P2'. Something is bigger than some planet.

Also easy, right? Now, from less formal to more formal.

S2. B(x',Py'), or even more formally...
S2'. ∃x',y' : B(x',y') & Py', or more compactly...
S2''. ∃x'y' Bx'Py', which is also dense.

Note that now y' is bound to the set P and is in a binary relation with unbound x'.

Third Part: x'' ⊃ y''

The statement 'everything a is b' is an idiom for 'if x then y', right? Now connect the two parts. Let x'' be the first part, and y'' be the second part, and we have:

S3: [∃x,y : Mx & O(x,y)] ⊃ [∃x',y' : B(x',y') & Py']

Oh boy, but we have four variables instead of two? How do they relate? Well, by the material implication we know that the unbound variable in the hypothesis is the unbound variable in the conclusion. That is, y must be x'. Let's use 'z' and replace 'y' and 'x*', and drop the 'y*' in favor of y.

S4: [∃x,z : Mx & O(x,z)] ⊃ [∃z,y : B(z,y) & Py]

Let's do a check-in natural language:

P3: Some moon x orbits around something z, and that implies that the something z is bigger than some planet y.

Is that the same as Q1 (Everything with a moon orbiting around it is bigger than a planet.)? Nope! The statement 'everything a is b' is an idiom for 'if a then b' only when it applies to ALL things in relation to both the a's and the b's. Let's add the universal quantifier.

S4: [∃z∃x : Mx & O(x,z)] ⊃ [∃z∃y : B(z,y) & Py], and since the 'z' is the same...
S5: ∃z[ ∃x Mx&O(x,z) ⊃ ∃y B(z,y)&Py ], but now let's make it all somethings and all planets...
S6: ∀z [ ∃x Mx&O(x,z) ⊃ ∀y B(z,y)&Py ]

Another natural language check:

P4: All z where some moon orbits z is bigger than all planets. Or in other words...
P5: For every something with a moon that orbits it, that something must be bigger than a planet.

How does that line up with Q1 (Everything with a moon orbiting around it is bigger than a planet.)? Well, by intuition, these two sentences contain the same deep structure, so we have our answer. (Unless I botched something because I'm more interested in my sandwich than the answer.)

Personally I like to bend notational convention and would try for the most dense expression possible:

S6: ∃Mx∀z,Py OMxz⊃BzPy or...
S7: ∃Mm∀xPp BOMmxPp which is quite opaque!

P6: There exists some moon for all somethings and all planets such that that moon which orbits the thing in question determines the thing in question must be bigger than all planets.

But now I'm just off in idiosyncratic territory generalizing quantifiers.