Does de morgans law apply to literals within brackets? This is the example I have, however, in the second line de morgans law does not turn the literals x,z, and y into not x,z,y. Why is this? Does deMorgans law not apply to those literals for some special reason?
≡ ∃x, y, ¬(¬(x < y) ∨ ∃z (x < z ∧ z < y)) (by p ⇒ q ≡ ¬p ∨ q)
≡ ∃x, y, ((x < y) ∧ ∀z (x ≥ z ∨ z ≥ y)) (byDeMorgan’s law)

In the language of my first order logic text, $x,y,z$ are individual variables, not propositional variables.  They don't represent truth values, they are merely arguments for the predicate variables that do return truth values.  Perhaps it's clearer to write the sentence without the infix inequality symbols: 
$$\exists x,y,\neg(\neg Lxy\vee\exists z(Lxz\wedge Lzy))$$
From here, it's much clearer to see what De Morgan's laws should and shouldn't do. 
$$\exists x,y,Lxy\wedge\neg\exists x(Lxz\wedge Lzy)$$
$$\exists x,y,Lxy\wedge\forall z\neg(Lxz\wedge Lzy)$$
$$\exists x,y,Lxy\wedge\forall z(\neg Lxz\vee\neg Lzy)$$
$$\exists x,y,Lxy\wedge\forall z(GExz\vee GEzy)$$
