# predicate first order logic satisfiability

I have problems to understand how to think of satisfiability in predicate logic.

For example why is $$\forall x (p(x) \vee \neg p(x))$$ valid, but

$$\forall x p(x) \vee \forall x\neg p(x)$$ only satisfiable?

And $$( \forall x (p(f(x)) \rightarrow \neg p(f(x)))) \land p(c)$$ satisfiable, but

$$( \forall x (p(f(x)) \rightarrow \neg p(f(x)))) \land p(f(c))$$ not?

$$c$$ is a constant, $$p$$ a predicate and $$f$$ a function.

It will help to rewrite the relevant expressions in natural language. Let's look at the first two. The first says "For every $$x$$, either $$p(x)$$ holds or $$p(x)$$ fails." This should obviously be true, without any other information: everything is either true or false, so for any specific object $$x$$ and any specific property $$p$$ either $$x$$ has $$p$$ or $$x$$ doesn't have $$p$$.
The second, however, says "Either every $$x$$ has property $$p$$ or every $$x$$ fails to have property $$p$$." Basically, this is saying that all objects look the same (at least as far as property $$p$$ is concerned). And this is clearly not always true: e.g. in the natural numbers, take $$p$$ to be evenness: some numbers are even, so $$\forall x(\neg p(x))$$ fails, but some numbers are odd, so $$\forall x(p(x))$$ fails too.
I suspect a lot of the confusion is coming from "algebraic" intuition - specifically in the case of the first two, the idea that everything is distributive: that we can distribute the "$$\forall$$" over the "$$\vee$$." There are indeed various "algebraic" rules, but they're not necessarily the ones you'll think of immediately; before you have a solid understanding of them, you need to resist the temptation to "simplify in the obvious way."