Вопрос не такой уж сложный на самом деле.
Импликация раскрывается так: A -> B = !A V B (!A - это НЕ А)
Заменяем каждую скобку
1) (!x1 V x2)(!x2 V x3)(!x3 V x4)(!x4 V x5) =
= (!x1*!x2 V x2*!x2 V !x1*x3 V x2*x3)*(!x3*!x4 V x4*!x4 V !x3*x5 V x4*x5) =
= (!x1*!x2 V !x1*x3 V x2*x3)*(!x3*!x4 V !x3*x5 V x4*x5) =
= !x1*!x2*!x3*!x4 V !x1*x3*!x3*!x4 V x2*x3*!x3*!x4 V !x1*!x2*!x3*x5 V
V !x1*x3*!x3*x5 V x2*x3*!x3*x5 V !x1*!x2*x4*x5 V !x1*x3*x4*x5 V x2*x3*x4*x5 =
= !x1*!x2*!x3*!x4 V !x1*!x2*!x3*x5 V !x1*!x2*x4*x5 V !x1*x3*x4*x5 V x2*x3*x4*x5 =
= !x1*!x2*!x3*(!x4 V x5) V !x1*x4*x5*(!x2 V x3) V x2*x3*x4*x5 = 1
2) Тоже самое с y
(!y1 V y2)(!y2 V y3)(!y3 V y4)(!y4 V y5) =
!y1*!y2*!y3*(!y4 V y5) V !y1*y4*y5*(!y2 V y3) V y2*y3*y4*y5 = 1
3) x1*y1 = 1, это значит, что x1 = y1 = 1, !x1 = !y1 = 0
Подставляем
!x1*!x2*!x3*(!x4 V x5) V !x1*x4*x5*(!x2 V x3) V x2*x3*x4*x5 = 1
0 V 0 V x2*x3*x4*x5 = 1
x2 = x3 = x4 = x5 = 1
!y1*!y2*!y3*(!y4 V y5) V !y1*y4*y5*(!y2 V y3) V y2*y3*y4*y5 = 1
0 V 0 V y2*y3*y4*y5 = 1
y2 = y3 = y4 = y5 = 1