Show that `(pvvq) to r -= (p to r) ^^ (q to r)`

Explanation

Truth table for `(pvvq) to r ` and `(p to r) ^^ ( q to r )`.

It is evident from the last two columns that
`(pvvq) to r -= ( pto r) ^^ ( q to r)`
Note:
`1 p to q -= (~p) vvq`
`2. p harr q -= (p to q) ^^ ( qto p)`
`3. p harr q -= (~ p vvq) ^^(pvv~q)`