【正确答案】(1)用等值演算法：
(P→(P∧Q))∨R=(￢P∨(P∧Q))∨R=((￢P∨P)∧(￢P∨Q))∨R=￢P∨Q∨R．
当P，Q，R的指派分别取不同真假值时，此公式的结果有时为真，有时为假，故是可满足式．
(2)真值表法：
真值表中结果有真有假，故是可满足式．
| P | Q | R | P∧Q | P→(P∧Q) | (P→(P∧Q))V R |
| F | F | F | F | T | T |
| F | F | T | F | T | T |
| F | T | F | F | T | T |
| F | T | T | F | T | T |
| T | F | F | F | F | F |
| T | F | T | F | F | rr |
| T | T | F | T | T | T |
| T | T | T | T | T | T |