English Sentence to Predicate Logic

Question

I have 2 English sentences that I am having trouble translating them to predicate logic.

1. All people who live together drive the same car.

2. There is no number that is larger than all other numbers.

I do know that for sentence 1 it has to have a universal quantifier. The key word that gave it was All. Would this be correct or close to the answer ..

∀x(Person(x) ∧ Home(x) -> ∃y Car(y,x))

For the second one Same thing... Instead of a universal i use existential for the first part because it only refers to 1 number. Can anyone help out with these? I'm just confused on how to tackle these. I'm looking for the answer but as well as the thoughts on how i can figure these out.

Answer 1

To simplify notation, use a two-sorted logic where lowercase letters are person variables and uppercase letters are car variables. Let $x\mathrm{T}y$ be the relation of $x$ and $y$ living together and $x\mathrm{D}A$ be the relation that $x$ drives $A$.

"All people who live together drive the same car." is equivalent to "For each person, there is a car that all people who live with him also drive." This is logically expressed as

$$ \forall x \exists A[x\mathrm{D}A \land \forall y[x\mathrm{T}y\rightarrow y\mathrm{D}A]]. $$

I'm not sure if it's intended that this car is the only car any of them drive, in which case the statement is $$ \forall x \exists A[x\mathrm{D}A \land \forall B[x\mathrm{D}B\rightarrow B=A]\land \forall y[x\mathrm{T}y\rightarrow (y\mathrm{D}A\land \forall C[y\mathrm{D}C\rightarrow C = A])]], $$ where $x\mathrm{D}A\land\forall B[x\mathrm{D}B\rightarrow B=A]$ expresses the idea that "the only car that $x$ drives is $A$".

Answer 2

I'd translate the first sentence like this: $$\begin{align}&\forall x\forall y \text{Person}(x)\wedge\text{Person}(y)\\ &\to\Big[(\exists h \wedge \text{Home}(h)\wedge\text{Live}(h,x)\wedge\text{Live}(h,y)) \\&\to\forall c\forall d\Big(\big(\text{Car}(c)\wedge\text{Car}(d)\big)\to[(\text{Drive}(c,x) \longleftrightarrow\text{Drive}(c,y))\\ &\wedge (\text{Drive}(c,x)\wedge\text{Drive}(d,x)\to c=d)]\Big)\Big]\end{align}$$

The second:

$$\neg\exists x\forall y(\text{Number}(x) \wedge \text{Number}(y)\to x>y)$$

Answer 3

For 1, try

$$\forall x,y(((\operatorname{Person}(x)\land \operatorname{Person}(y))\land\operatorname{LivesTogether}(x,y))\to(\exists z\operatorname{Car}(z)\land(\operatorname{Drives}(x,z)\land\operatorname{Drives}(y,z)))$$

and for 2, try

$$\lnot\exists n\forall m(n>m),$$

where the domain for the second one is understood to mean all (real) numbers.