You can
put this solution on YOUR website! (S'U T')'
DeMorgan's Laws are:
1. (A ᑌ B)' = A' ᑎ B'
and
2. (A ᑎ B)' = A' ᑌ B'
To do your problem we need the first one:
(A ᑌ B)' = A' ᑎ B'
Substitute (S') for A
Substitute (T') for B
Substitute [ for (
Substitute ] for )
[(S') ᑌ (T')]'
becomes
(S')' ᑎ (T')'
Substitute S for (S')' and T for (T')'
S ᑎ T
Edwin