Correct Answer - D
`(p wedge q) to (~p) vee q`
`-= (~(p wedge q)) vee ((~p) vee q)`
` -= ((~p) vee (~q)) vee ((~p) vee q)`
`-=(~p) vee (~q) vee q " "[ because (~p) vee (~p) -= ~p]`
` -= (~p) vee T " "[because ~q vee q -= T]`
`-= T`
So, it is a tautology `" " [ because (( ~q) vee q) " is tautology"]`
(b) `(p wedge q) to p -= (p wedge q) vee p`
`-= (( ~p) vee (~q)) vee p`
` " "[ because ~ (p wedge q) -= (~p) vee (~q)]`
`-= (~ p vee p ) vee (~q)` is tautology.
` [ because ~ p vee p " is a tautology and " (~q) vee T -= T]`
(c) `because p to (p vee q) -= (~p) vee (p vee q)`
`[ because p to q " is equivalent and " (~p vee q)]`
`-= (~p vee p ) vee q` is tautology.
`[ because (~p vee p) " is tautology and " q vee T -= T]`
(d) `(p vee q) to (p vee (~q))`
`-= (~(p vee q)) vee (p vee (~q))`
`-= ((~p) wedge (~q)) vee (p vee (~q))`
`-= (p vee (~q) vee ((~p ) wedge (~q))`
`-= (p vee (~q) vee (~p)) wedge (p vee (~q) vee (~q))`
`-= (T vee (~q)) wedge (p wedge (~q))`
`-= T wedge (p wedge (~q))`
`-= p wedge (~q),` which is not a tautology.
