When to use conjunction and when to use implication? (First Order Logic) [closed]

Question

I'm learning First Order Logic at the moment. I'm looking at this example:

Some dogs bark ∃x (dog(X) Λ bark(x))

All dogs have four legs ∀x (dog(x) -> have_four_legs(x))

My question is: is it possible for the second example to be: ∀x (dog(x) Λ have_four_legs(x))

And why can't the first example be: ∃x (dog(X) -> bark(x))

Overstrung

btw, this would be better posted at cstheory.stackexchange.com – Adown

This question appears to be off-topic because it is about mathematical logic, which is more appropriate at math.stackexchange.com. – Bearnard

whether this is offtopic or ontopic, this is valuable question :) – Carin

Answer 1

Well ask yourself this: Are implication and conjunction equivalent? No. Your last statement says that all x's are both dogs and have four legs. While that does mean that all dogs have four legs, it also means that everything is a dog...

I suggest writing out what each statement means in English:

There is some x that is a dog and barks There is some x where if it is a dog it barks

Now can you see the differences? The second one doesn't specifically say that a dog exists.

Adown

ah I see, so what about: ∃x (dog(X) -> bark(x)) . Wouldn't that be valid? – Overstrung

see edits...I should add that I havn't done this for a year or so, but thinking logically normally works – Adown

Answer 2

∃x (dog(X) -> bark(x))

Late reply, but if anyone does end here and wants to know, from what i've been learning that means:

There exists a dog that barks vs some dogs bark.

More precise:

there exists some x, if x is a dog, then it barks. -> is an if-then statement.

∃x (dog(X) Λ bark(x)) means there exists some dog and it barks, in other words, some dogs bark.

∀x (dog(x) Λ have_four_legs(x)): Everything is a dog AND everything has 4 legs.

∀x (dog(x) -> have_four_legs(x)) for everything IF dog THEN it has 4 legs.

Pryer