Recent content by RonnieRongz

[(AB')+(A'B)]' By De Morgan's Theorem (AB')' • (A'B)' Again by De Morgan's Theorem (A'+B'') • (A''+B') Simplify (A'+B) • (A+B') Multiply A'A+A'B'+AB+BB' *AA' and BB' are equal to 0 Therefore [(AB')+(A'B)]' = A'B'+AB Hope this helps