Question

数电中这个代数是怎么化简的？

Answer 1

Y = ABC+(A+B+C)(BC+AC+AB)'
= ABC+(A+B+C)[(AB)'(BC)'(AC)']
= ABC+(A+B+C)[(A'+B')(B'+C')(A'+C')]
= ABC+(A+B+C)[(A'B'+A'C'+B'+B'C')(A'+C')]
= ABC+(A+B+C)(A'C'+B')(A'+C')
= ABC+(A+B+C)(A'C'+A'B'+B'C')
= ABC+(AB'C'+A'BC'+A'B'C)
另外：
A ⊕ B ⊕ C = (A ⊕ B) ⊕ C = (AB'+A'B) ⊕ C = (AB'+A'B)C' + (AB'+A'B)'C
= AB'C'+A'BC' + [(AB')'(A'B)']C = AB'C'+A'BC' + [(A'+B')(A+B')]C
= ABC + (AB'C'+A'BC'+A'B'C)
从而证明： Y = A ⊕ B ⊕ C.

Answer 2

3输入............................上图，
Y的第一项。