Everyone loves a lover

Question

I need help translating this into predicate logic and solving this proof:

Premise 1: Everyone loves a lover

Premise 2: Shelley is a lover

Conclusion: So, everyone loves everyone

-I translated the first premise as (Az)(Ax)(Ay)(Lxy --> Lzx)

with A being the universal quantifier

-I translated the second premise as (Ey)Lsy

with E being the existential quantifier

-I translated the conclusion as (Ax)(Ay)Lxy

with A being the universal quantifier

My professor said that my translations worked but now I still need help with the proof. Can anyone help?

Answer 1

You want to define all of the symbols that you introduce in your proof. For example:

Let s be the person "Shelley" and let L(x,y) be a predicate meaning person x loves person y.

There is some ambiguity in the problem statement so I'll assume person x is a lover if there exists a person y such that L(x,y). The first premise also could be reasonably interpreted as either a universal statement (everybody loves every lover) or an existential statement (everybody loves at least one lover), but only the latter will prove the conclusion.

To get started on the proof you have

1. ∀x∀y∀z(L(x,y)→L(z,x)) (Premise 1)
2. ∃x(L(s,x)) (Premise 2)
3. L(s,a) (∃-elimination on 2)
4. L(s,a)→L(z,s) (∀-elimination on 1)
5. L(z,s) (3,4 Modus Ponens)
6. L(z,s)→L(w,z) (∀-elimination on 1)
7. L(w,z) (5,6 Modus Ponens)∀
8. ∀x∀y(L(x,y)) (∀-introduction on 7)

The key points to review here would probably be the rules for quantifier introduction and elimination.

From a universal statement we can substitute an element of the domain or an arbitrary placeholder (in this proof only s is a real domain element, the rest of the variables used were arbitrary). From an existential statement we can substitute a placeholder as long as it has not been earlier in the proof.

From any predicate P(x) and argument we can introduce the existential statement ∃aP(a) and if the argument is an arbitrary variable we can introduce the universal statement ∀xP(x)