(10) Draw a truth table for the compound statementq∨p→((p∧q)∨(q∧r))" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap;...

Solution 1: q∨p→((p∧q)∨(q∧r))
                    Truth Table
    p
    q
    r
    p∧q
    q∧r
    q∨p
    (p∧q)∨(q∧r)
    q∨p→((p∧q)∨(q∧r))
    T
    T
    T
    T
    T
    T
    T
    T
    T
    T
    F
    T
    F
    T
    T
    T
    T
    F
    T
    F
    F
    T
    F
    F
    T
    F
    F
    F
    F
    T
    F
    F
    F
    T
    T
    F
    T
    T
    T
    T
    F
    T
    F
    F
    F
    T
    F
    F
    F
    F
    T
    F
    F
    F
    F
    T
    F
    F
    F
    F
    F
    F
    F
    T
Solution 2: ((s∨r∨¬q)∧(¬s∨¬r∨s)∧(r∨¬s∨¬q))=(¬q∨r)
Considering LHS,
(1) ((s∨r∨¬q)∧(¬s∨¬r∨s)∧(r∨¬s∨¬q)),
Using commutativity, ¬r∨s =s∨¬r,
(s∨r∨¬q)∧(¬s∨s∨¬r)∧(r∨¬s∨¬q))
Because ¬s∨s = T, Therefore, LHS reduces to,
(s∨r∨¬q)∧(T∨¬r)∧(r∨¬s∨¬q))
Now, using Domination Law, (T∨¬r) = T,
(s∨r∨¬q)∧(T)∧(r∨¬s∨¬q)),
Using Identity law, (s∨r∨¬q)∧(T) = (s∨r∨¬q). Hence,
(s∨r∨¬q)∧(r∨¬s∨¬q))
By Distributive law,
r∨((s∨¬q)∧(¬s∨¬q)) =...