# ∀x.Q(x) -> Q(a), is it true or false? Why?

∑ = ( { 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...

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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

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.
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
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.