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A1:[P→(Q→R),Q]⟹(P→R)

=> [!P V !Q V R, Q]

=> !P V R

=> (P->R)

A2:[(P→Q),(~Q→R),~R]⟹P

By Modus Tollens, Q is conclusion. P can be 0/1 when Q is 1. So invalid.

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