Negation plus the rules of propositional calculus

[solakis1]

Prove the following:

\(\displaystyle \neg\forall\epsilon[\epsilon>0\Longrightarrow\exists\delta(\delta>0\wedge\forall x(x\in Df\wedge 0<|x-a|<\delta\Longrightarrow |f(x)-l|<\epsilon))]\)\(\displaystyle \Longleftrightarrow\exists\epsilon[\epsilon>0\wedge\forall\delta(\delta>0\Longrightarrow\exists x(x\in Df\wedge 0<|x-a|<\delta\wedge |f(x)-l|\geq\epsilon))]\)

By using the 4 basic rules of predicate calculus plus the rules of propositional calculus

 

[Valkarie]

, the proof can be established as follows:1. $\neg\forall\epsilon[\epsilon>0\Longrightarrow\exists\delta(\delta>0\wedge\forall x(x\in Df\wedge 0<|x-a|<\delta\Longrightarrow |f(x)-l|<\epsilon))] \equiv \exists\epsilon[\neg(\epsilon>0\Longrightarrow\exists\delta(\delta>0\wedge\forall x(x\in Df\wedge 0<|x-a|<\delta\Longrightarrow |f(x)-l|<\epsilon)))]$ (Negation of Universal Quantifier)2. $\exists\epsilon[\neg(\epsilon>0\Longrightarrow\exists\delta(\delta>0\wedge\forall x(x\in Df\wedge 0<|x-a|<\delta\Longrightarrow |f(x)-l|<\epsilon)))] \equiv \exists\epsilon[\epsilon>0\wedge\neg(\exists\delta(\delta>0\wedge\forall x(x\in Df\wedge 0<|x-a|<\delta\Longrightarrow |f(x)-l|<\epsilon)))]$ (De Morgan's Law)3. $\exists\epsilon[\epsilon>0\wedge\neg(\exists\delta(\delta>0\wedge\forall x(x\in Df\wedge 0<|x-a|<\delta\Longrightarrow |f(x)-l|<\epsilon)))] \equiv \exists\epsilon[\epsilon>0\wedge\forall\delta(\neg(\delta>0\wedge\forall x(x\in Df\wedge 0<|x-a|<\delta\Longrightarrow |f(x)-l|<\epsilon))