(((P→Q)∨S)↔T) == VALID
((P→(Q∧(S→T)))) == INVALID
(¬(B(¬Q))∧R) == INVALID
(P∧((¬Q)∧(¬(¬(Q↔(¬R))))))  ==  VALID
((P∨Q)→¬(P∨Q))∧(P∨(¬(¬Q))) == INVALID

Solved using ex. 2