Existential and universal Quantifiers in AI

Question

I have just started first order predicate logic. Why does universal quantifier and single implication go together? Similarly existential quantifier and conjunction go together?

Take the statement: Some frogs are green Why then is this an incorrect translation: ∃x (frog(x) → green(x))?

Also, for the statement: All frogs are green (∀x)(frog(x) → green(x)) does not seem a correct translation; as in the cases where frog(x) becomes false the expression (∀x)(frog(x) → green(x)) will be always true.

Explaining with truth table will be helpful

Answer 1

I suppose that by "go together" you mean the following:

All frogs are green means the same as for all things, if they are a frog, they are green, or, as a formula of first order predicate logic:

(∀x)(frog(x) → green(x))

Some frogs are green means the same as some things are both frog and green i.e.

(∃x)(frog(x) ∧ green(x))

Then you ask:

Take the statement: Some frogs are green Why then is this an incorrent translation: (∃x)(frog(x) → green(x))

Now, if some frogs are green, indeed (∃x)(frog(x) → green(x)) is true! But the converse is not true: As soon as there is one non-frog in your domain, (∃x)(frog(x) → green(x)) is true, even if there is no green frog to be seen, ie. even if Some frogs are green is false.

So (∃x)(frog(x) → green(x)) doesn't mean the same as some frogs are green

(see also this discussion at philosophy.stackexchange.com)

Also, for the statement: All frogs are green (∀x)(frog(x) → green(x)) does not seem a correct translation; as in the cases where frog(x) becomes false the expression (∀x)(frog(x) → green(x)) will be always true.

I think you want to say here that if there are no frogs (∀x)(frog(x) → green(x)) is true, while in your (and Aristotle's) opinion All frogs are green is not, in that case. This is the old and venerable question of the existential import of universal propositions. It has a long and fascinating history; suffice to say that modern logicians and philosophers hold that in a world without frogs All frogs are green is true (as is All frogs are red)

Answer 2

If some frogs are green, then: ∃ x : Frog(x) => Green(x) Sounds correct to me...