AB!AB+B = B, { т.к. A&!A = 0 }
(A+B)(!A+!B)=A!A+A!B+B!A+B!B = 0 + A!B+B!A + 0 = A&!B + B&!A
A+AB+AC = A(1+B+C)=A&1 = A
A+!AB+!AC = A+!A(B+C)
A(A+B+C)=AA+AB+AC=A+AB+AC=A(1+B+C)=A&1 = A
AB+!B+!AB =B(A+!A)+!B=B+!B = 1
(!A+B)!C(C+A!B) = !A!CC+!A!CA!B+B!CC+B!CA!B=0+0+B!C+B!CA!B =
!C(B+BA!B)=!C(B+0)=!CB
A(!B!C+BC)+A(B!C+!BC) = A!B!C+ABC+AB!C+A!BC =
A(!B!C+BC+B!C+!BC) = A(!B(!C+C)+B(C+!C)) = A(!B+B)=A
{т.к. (x+!x)=1}
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A!(!B+C)=A&(B&!C) = AB!C
{т.к. !(x+y)=!x&!y }
!(A+!B)+!(A+B)+AB = !AB+!A!B+AB = !A(B+!B)+AB = !A+AB
A+!(A+B)+!AB = A+!A!B+!AB = A+!A(!B+B)=A+!A = 1
!(A+!B+!C) = !ABC
!(A+B)A!B = !AA!B + BA!B = 0 + 0 = 0
A+!(B!C)+!(!A+B+!C) = A+!B+C+A!BC = A(1+!BC)+C = A+C
(A+B+C)!(!A!B)+C=(A+B+C)(A+B)+C = AA+AB+BA+BB+CA+CB+C =
= A+BA+B+CA+CB+C = A(1+B)+B+C(A+B+1)=A+B+C
A!(!C+!B)+!(!A+B)C+AC = ACB+A!BC+AC = AC(B+!B+1)=AC
(A+B)(!A+B)(!A+!B)=(A!A+AB+B!A+BB)(!A+!B) =
= (0+B(A+!A+1))(!A+!B) = B(!A+!B) = B!A+B!B = B!A