① A? ((C∧? D)∨(? C∧D)) A∨(C∧? D)∨(? C∧D)
② ? (B∧C) B∨? C
③ CDC ∨? D
The general situation is: formula (? A∨(C∧? D)∨(? C∧D))∧(? B∨? C)∧(? C∨? D) correct.
(? A∨(C∧? D)∨(? C∧D) )∧(? B∨? C)∧(? C∨? d)
(? A∨(C∧? D)∨(? C∧D) )∧(? C∨(? B∧? d))
(? A∧? C)∨(C∧? D∧? C)∨(? C∧D∧? C)∨(? A∧? B∧? D)∨(C∧? D∧? B∧? D) ∨(? C∧D∧? B∧? d)
(? A∧? C)∨F∨(? C∧D)∨(? A∧? B∧? D)∨(C∧? D∧? b)∞F
(? A∧? C)∨(? C∧D)∨(? A∧? B∧? D)∨(C∧? D∧? B) obtaining the disjunctive normal form of the formula
To make the original formula hold, then (? A∧? c),(? C∧D),(? A∧? B∧? d),(C∧? D∧? At least one of b) is true. According to the condition of "two out of four people are on business trip", the possible sending methods are as follows:
Take? A∧? C is t, which means b and d go.
Take? C∧D is T, which means A and D go, or B and D go.
Get a C∧? D∧? B is t, which means a and c go.
Finally, there are three sending methods: A and C, A and D, B and D. ..