1) Раскрываем скобки, как в алгебре
F = (A V B V C) /\ (A V ~B V C) = (A /\ A) V (B /\ A) V (C /\ A) V
V (A /\ ~B) V (B /\ ~B) V (C /\ ~B) V (A /\ C) V (B /\ C) V (C /\ C) =
= A V (A /\ B) V (A /\ C) V (A /\ ~B) V 1 V (C /\ ~B) V (A /\ C) V
V (B /\ C) V C
По закону поглощения
A V (A /\ B) = A, поэтому
F = A V 1 V C = A V C
Таблица истинности
A | B | C | F
0 | 0 | 0 | 0
0 | 0 | 1 | 0
0 | 1 | 0 | 0
0 | 1 | 1 | 0
1 | 0 | 0 | 0
1 | 0 | 1 | 1
1 | 1 | 0 | 0
1 | 1 | 1 | 1
2) (0 V 1) V (1 V 0) = 1 V 1 = 1
((0 /\ 0) V 0) /\ (1 V 1) = (0 V 0) /\ 1 = 0 /\ 1 = 0
3) B -> (C V ~A)
Таблица истинности для импликации A -> B = F
A | B | F
0 | 0 | 1
0 | 1 | 1
1 | 0 | 0
1 | 1 | 1
В нашем случае
A | B | C |
C V ~A
| F
0 | 0 | 0 | __ 1 __ | 1
0 | 0 | 1 | __ 1 __ | 1
0 | 1 | 0 | __ 1 __ | 1
0 | 1 | 1 | __ 1 __ | 1
1 | 0 | 0 | __ 0 __ | 1
1 | 0 | 1 | __ 1 __ | 1
1 | 1 | 0 | __ 0 __ | 0
1 | 1 | 1 | __ 1 __ | 1
4) ~(X V Y) V ~(Y /\ Z) = (~X /\ ~Y) V ~Y V ~Z =
= (~X V ~Y V ~Z) /\ (~Y V ~Y V ~Z) = (~X V ~Y V ~Z) /\ (~Y V ~Z) =
= (~X /\ ~Y) V ~Y V (~Y /\ ~Z) V (~X /\ ~Z) V (~Y /\ ~Z) V ~Z = ~Y V ~Z
Логическую схему сами рисуйте, я в Пайнте это нарисовать не могу.
5) F = A /\ B