The goal of the following exercice is to show that $(($p $->$ q$)$ p$)->$ q is a tautology without using a truth table. At each step, identify what law was used to obtain the logical equivalence. $(($p $->$ q$)$ p$)->$ q $(($ p $$ q$)$ p$)->$ q By $($i$)(($ p $$ q$)$ p$)$ q By $($ii$)(($ p $$ q$)$ p$)$ q By $($iii$)(($ p$)$ q$)$ p$)$ q By $($iv$)(($p$$ q$)$ p$)$ q By $($v$)($ p $($p$$ q$))$ q By $($vi$)(($ p $$ p$)($ p$$ q$))$ q By $($vii$)($T $($ p$$ q$))$ q By $($viii$)($ p$$ q$)$ q By $($ix$)$ p $($ q $$ q$)$ By $($x$)$ p $$ T By $($xi$)$ T By $($xii$).$