If one expression is equivalent to a second expression, and the second

expression is equivalent to a third expression, then the first expression is

equivalent to the third expression. $($This is called "transitivity" of equivalence.$)$

Explain in words why the expressions $"P=>Q"$ and $"$ not$(not{Q}^{{?}^{?}}P)"$ are equivalent,

without using truth$-$tables: Use De Morgan's law, transitivity, and equivalences

that were proved in previous questions.

For each of the following expressions, write down an expression that is

equivalent to them but does not contain an implication $($i$.$e$.,$ no $"=>"$ symbol$)$:

I. $P=>(QvvR)$

II$.(P^{{?}^{?}}Q)=>R$

III. $P=>(Q=>R)$

IV$.(P^{{?}^{?}}Q)=>(R=>Q)$