1. [24 marks] Digital Logic. a) 'X' is the ternary connective such that 'Xpqr' is logically equivalent to '(p /\ q) \/ (q + r)'. '+' is 'xor'. 'F' and 'T' denote the 0-ary connectives 'false' and 'true', respectively. Without destroying correctness, put all 'sentence' letters in alphabetical order, and show only answers in which each 'sentence' letter is as far to the left as possible. Using {'X', 'F', 'T'}, synthesize: ~p |= =| X ____ ____ ____ Using {'X', '~', 'T'}, synthesize: p \/ q |= =| X ____ ____ ____ Using {'X', '~', 'T'}, synthesize: p /\ q |= =| ____ X ____ ____ ____ b) 'Y' is the ternary connective such that 'Ypqr' is logically equivalent to '(p <--> q) --> (q + r)'. '+' is 'xor'. p --> q |= =| ~p \/ q. 'p <--> q' is true iff 'p' and 'q' have the same truth value. 'F' and 'T' denote the two 0-ary connectives 'false' and 'true', respectively. Without destroying correctness, put all 'sentence' letters in alphabetical order, and show only answers in which each 'sentence' letter is as far to the left as possible. Using {'Y', 'F', 'T'}, synthesize: ~p |= =| Y ____ ____ ____ Using {'Y', 'F', 'T'}, synthesize: p \/ q |= =| Y ____ ____ ____ Using {'Y', '~', 'F', 'T'}, synthesize: p /\ q |= =| ___ Y ____ ____ ____