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187

answers:

2

Hi there,

I have a logical statement that says "If everyone plays the game, we will have fun".

In formal logic we can write this as:

Let D mean the people playing. Let G be the predicate for play the game. Let F be the predicate for having fun.

Thus [VxeD, G(x)] -> [VyeD, F(y)]

V is the computer science symbol for universal quantification. E below is the existential quantifier.

I'm looking for a way to write a similar statement using only existential quantifiers. My best guess would be that we simply need to find a way to find the counter-example where it doesn't happen, thus negate the above.

The problem is negating it doesn't make sense. We get:

[VxeD, G(x)] ^ [EyeD, !L(y)]

It's not a proper statement since the universal is still in there though it is also equivalent. Thus I need to re-fabricate my statement to something like: VxeD, VyeD, G(x) ^ F(y) I would get ExeD, EyeD, !G(x) v !F(y) which would mean "There exists someone who doesn't learn or someone else who doesn't have fun" which doesn't seem correct to me.

Some guidance or clarification would be fantastic :-)

Thanks!

+2  A: 

I don't understand your ^ symbol, but I believe you are looking for the contrapositive. In your example, if the original statement is:

[VxeD, G(x)] -> [VyeD, F(y)]

then the contrapositive is

[ExeD, !F(x)] -> [EyeD, !G(y)]

meaning "if there is someone who is not having fun, then there exists someone not playing the game." Note that this is different than the statement in your comment above: it may well be the case that everyone is having fun, but not everyone is playing.

In general, p -> q is equivalent to !q -> !p.

(Of course I may not have understood your notation correctly.)

elcelista
^ means "and" and v means "or" that's basically what I got in my answer. Great, you've affirmed what I had thought! Cheers.
+2  A: 

I'm having trouble reading your notation. I'll use A for the universal quantifier, E for the existence quantifier, F for the predicate 'having fun', G for the predicate 'playingng learned the game', then

AxL(x) -> AxF(x)

Now, you can just apply the usual gymnastics:

<==> !AxL(x) <- !AxF(x)

<==> Ex!G(x) <- Ex!F(x)

<==> Ex!F(x) -> Ex!G(x)

so, indeed, when someone's not having fun, it means not everybody played the game.

Much cleaner than the original.
Nerdling