用了异或门,2个与门,2个非门 ,或门
F=A(B'C+BC')+A'(B'C+BC')'
=AB'C+ABC'+A'(B'C)'(BC')'
=AB'C+ABC'+A'(B+C‘)(B’+C)
=AB'C+ABC'+A'(BC+B’C')
=AB'C+ABC'+A'BC+A'B’C'
到此最简了
用了异或门,2个与门,2个非门 ,或门
F=A(B'C+BC')+A'(B'C+BC')'
=AB'C+ABC'+A'(B'C)'(BC')'
=AB'C+ABC'+A'(B+C‘)(B’+C)
=AB'C+ABC'+A'(BC+B’C')
=AB'C+ABC'+A'BC+A'B’C'
到此最简了