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))

Was it helpful?

Solution

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.

OTHER TIPS

∃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.

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