Question

Информ. - тема Логика
Упростить следующие выражения:

 

1. (X v Z v Y) & (X v ¬Y v Z­­) & (X v ¬Y v ¬Z)

2. (A v ¬B) v A v F & E v B

3. X v T & Y & Z & ¬(X v T)

Answer 1

1)

 

(X+Z+Y)(X+!Y+Z)(X+!Y+!Z) -> X(X+!Y+Z)(X+!Y+!Z)+Z(X+!Y+Z)(X+!Y+!Z)+Y(X+!Y+Z)(X+!Y+!Z)
X(X+!Y+Z)(X+!Y+!Z) -> XX(X+!Y+!Z)+!YX(X+!Y+!Z)+ZX(X+!Y+!Z)
XX(X+!Y+!Z) -> XX+!YX+!ZX
XX -> X
XX+!YX+!ZX -> X
!YX(X+!Y+!Z) -> X!YX+!Y!YX+!Z!YX
X!YX -> X!Y
!Y!YX -> !YX
X!YX+!Y!YX+!Z!YX -> X!Y
ZX(X+!Y+!Z) -> XZX+!YZX+!ZZX
XZX -> XZ
!ZZX -> 0
XZX+!YZX+!ZZX -> XZ
XX(X+!Y+!Z)+!YX(X+!Y+!Z)+ZX(X+!Y+!Z) -> X
Z(X+!Y+Z)(X+!Y+!Z) -> XZ(X+!Y+!Z)+!YZ(X+!Y+!Z)+ZZ(X+!Y+!Z)
XZ(X+!Y+!Z) -> XXZ+!YXZ+!ZXZ
XXZ -> XZ
!ZXZ -> 0
XXZ+!YXZ+!ZXZ -> XZ
!YZ(X+!Y+!Z) -> X!YZ+!Y!YZ+!Z!YZ
!Y!YZ -> !YZ
!Z!YZ -> 0
X!YZ+!Y!YZ+!Z!YZ -> !YZ
ZZ(X+!Y+!Z) -> XZ+!YZ+!ZZ
!ZZ -> 0
XZ+!YZ+!ZZ -> XZ+!YZ
XZ(X+!Y+!Z)+!YZ(X+!Y+!Z)+ZZ(X+!Y+!Z) -> XZ+!YZ
Y(X+!Y+Z)(X+!Y+!Z) -> XY(X+!Y+!Z)+!YY(X+!Y+!Z)+ZY(X+!Y+!Z)
XY(X+!Y+!Z) -> XXY+!YXY+!ZXY
XXY -> XY
!YXY -> 0
XXY+!YXY+!ZXY -> XY
!YY(X+!Y+!Z) -> 0
ZY(X+!Y+!Z) -> XZY+!YZY+!ZZY
!YZY -> 0
!ZZY -> 0
XZY+!YZY+!ZZY -> XZY
XY(X+!Y+!Z)+!YY(X+!Y+!Z)+ZY(X+!Y+!Z) -> XY
X(X+!Y+Z)(X+!Y+!Z)+Z(X+!Y+Z)(X+!Y+!Z)+Y(X+!Y+Z)(X+!Y+!Z) -> X+!YZ

 

2)

(A+!B)+A+FE+B -> 1

 

3)

!(X+T) -> !X!T
TYZ(!(X+T)) -> 0
X+TYZ(!(X+T)) -> X