$3$ Negation and Conjunction

Show that the Law of Excluded Middle implies one of DeMorgan$$s Laws, $($P :Prop,$$P $$P$)=($PQ:Prop,$($P $$Q$)=$P $$Q$).$ following the instructions for an acceptable proof. The Coq statement is

Lemma prob$3$ : $($forall P : Prop, ~ P $\backslash /$ P$)->($forall P Q : Prop, ~ $($P $/\backslash$ Q$)->$ ~ P $\backslash /$ ~ Q$)$

Use the template:

Lemma prob$3$ : $($forall P : Prop, ~ P $\backslash /$ P$)->($forall P Q : Prop, ~ $($P $/\backslash$ Q$)->$ ~ P $\backslash /$ ~ Q$).$

Proof.

Qed.