P→[P∧(q→P)]=[P∧(q→P)]∨!P=(P∧(P∨!q))∨!P=(p∨!P)∧(P∨!q∨!P)=true
[(p→r)∨q]∧(q→r)=((r∨!p)∨q)∧(r∨!q)=(r∨!q)∧(!p∨r∨!q)∧(q∨r∨!q)
=r∨!q
!means not
P→[P∧(q→P)]=[P∧(q→P)]∨!P=(P∧(P∨!q))∨!P=(p∨!P)∧(P∨!q∨!P)=true
[(p→r)∨q]∧(q→r)=((r∨!p)∨q)∧(r∨!q)=(r∨!q)∧(!p∨r∨!q)∧(q∨r∨!q)
=r∨!q
!means not