Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
∑ = ( { a, b, c, d }, { f¹ }, { P², Q¹ } )
U = { ♡, ♢, ♣, ♤ }

     { a -> ♤, b -> ♡, c -> ♣, d -> ♤ },
f∑ = { f¹ -> { ♤ -> ♣, ♣ -> ♤, ♡ -> ♢, ♢ -> ♡ } },
     { P² -> { (♡, ♣), (♡, ♤), (♢, ♣), (♢, ♤) }, Q¹ = { ♢, ♡ } }

∀x.Q(x) -> Q(a) is it true or false? I've been told that it is true, but

Q(b) -> Q(a)
T    -> F        -- should be false, right?

Am I missing something here?

It is written exactly as ∀x.Q(x) -> Q(a), no extra parenthesis or anything...

share|improve this question
9  
Yes, you're missing something here. The first thing you're missing is one or more complete sentences that explain what the heck you're on about, rather just some unexplained lines of symbols. –  Henning Makholm Oct 8 '12 at 18:09

2 Answers 2

up vote 1 down vote accepted

One needs to use parentheses properly, or interpret the arguably ambiguous expression as intended. It looks as if the dot is intended to bind the $\forall x$ to $Q(x)$. So I would interpret the expression as $(\forall x Q(x))\rightarrow Q(a)$. An analysis along your lines will show that it is true.

As for the less plausible interpretation $\forall x(Q(x)\rightarrow Q(a))$, that is false in most structures.

share|improve this answer
    
Could you explain (∀xQ(x))→Q(a) a little bit? I don't understand how it is true... –  BrunoLM Oct 8 '12 at 18:24
    
In your structure (that is, with your interpretation of $Q$ and $a$) the antecedent $\forall x Q(x)$ is false, so the implication is true. In any structure, if (the interpretation of) $Q$ holds at every object, then it holds in particular at (the interpretation of) $a$. –  André Nicolas Oct 8 '12 at 18:28
    
Oh, now I see it, I believe I was reading as the second interpretation... Thanks a lot! –  BrunoLM Oct 8 '12 at 18:31

Pleas check carefully: Do you mean $$\forall x\colon (Q(x)\to Q(a))$$ or $$(\forall x\colon Q(x))\to Q(a)$$ ? That should make things clear.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.