Dual of `(x vv y) ^^ (x' vv 1)` is

Dual of (x vv y) ^^ (x vv 1) = x vv x ^^ y vv y is

Dual of (x +y) (x'*1) is

Dual of x ^^(y vv x)=x is

Observe the following statements I : The dual of [ ~p ^^ q)] vv [ p ^^ {~(q vv ~s)}] is [~(p vv q)] ^^ [p vv{~(q ^^ ~s)}] II : The dual of ~p ^^ [(~q) ^^ (p vv q) ^^ ~r] is ~p vv[~q) vv(p ^^ q) vv ~r] The true statements in the above is/are :

Observe the following statements I : The dual of [ ~p ^^ q)] vv [ p ^^ {~(q vv ~s)}] is [~(p vv q)] ^^ [p vv{~(q ^^ ~s)}] II : The dual of ~p ^^ [(~q) ^^ (p vv q) ^^ ~r] is ~p vv[~q) vv(p ^^ q) vv ~r] The true statements in the above is/are :

The dual of the statement (p vv q) ^^ ~(r vv s) is

Simplify (p vv q) ^^ (p vv ~q) .