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Home/ Questions/Q 367105
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T13:41:47+00:00 2026-05-12T13:41:47+00:00

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

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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!

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  1. Editorial Team
    Editorial Team
    2026-05-12T13:41:47+00:00Added an answer on May 12, 2026 at 1:41 pm

    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.)

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