Prove that `q to p -= ~p to ~q`

Prove that ~(~pto ~q) -=~p ^^q

Prove that p to (~q vee r) -= ~p vee (~q vee r) using truth table.

Prove p to (q to r) -= (p wedge q) to r without using truth table.

Prove that a^p b^ q <((a p+b q)/(p+q))^(p+q)dot

Prove that ~((~p)^^q) -=pvv(~q) .

If the p^("th"), q^("th") and r^("th") terms of a G.P. are a, b and c, respectively. Prove that a^(q – r) b^(r-p) c^(P - q) = 1.

If lines p x+q y+r=0,q x+r y+p=0a n dr x+p y+q=0 are concurrent, then prove that p+q+r=0(w h e r ep ,q ,r are distinct )dot

Prove that p + q = 0 " if " x^(3) - 3px + 2q " is divisible by" (x + 1)^(2)

If a and b are the roots of x^2 – 3x + p = 0 and c, d are roots of x^(2) – 12x + q = 0 , where a,b,c,d form a G.P Prove that (q+p):(q-p) = 17 : 15

The following statement (p to q) to [(~p to q) to q] is