1. [20 marks] Digital Logic. If a task is impossible, write "impossible". a) '#' is the ternary 'majority' connective. '#pqr' is true iff a majority of 'p', 'q', and 'r', is true. 'F' is the 0-ary 'false' connective. 'T' is the 0-ary 'true' connective. Using only {'#', 'F', 'T'}, complete: ~p |= =| # ___ ___ ___ Using only {'#', 'F', 'T'}, complete: p /\ q |= =| # ___ ___ ___ b) Let 'A' be the ternary connective such that 'Apqr' is equivalent to '(p --> q) \/ (p --> r)'. We have: p --> q |= =| ~p \/ q. 'F' is the 0-ary 'false' connective. 'T' is the 0-ary 'true' connective. Using only {'A', 'F'}, complete: ~p |= =| A ___ ___ ___ Using only {'A', '~', 'T'}, complete: p /\ q |= =| ___ A ___ ___ ___