The statement which is not a tautology is `(pharrq)[(pvecq)^^(qvecq)]` b. `(pvecq)vv(~ pvecq)` c. `(p^^q)vec(pvvq)` d. `(p^^q)vv[(pvvq)vec~ q]`

The statement which is not a tautology is (p harr q)[(p rarr q)^^(q rarr q)] b.(p rarr q)vv(sim p rarr q) c.(p^^q)rarr(p vv q)d(p^^q)vv[(p vv q)rarr-q]

The statement ~(p ^^ q) vv q

The statement ~(p ^^ q) vv q

The statement pvec(qvecp) is equivalent to (1) pvec(pvecq) (2) pvec(pvvq) (3) pvec(p^^q) (4) pvec(pharrq)

The logically equivalent proposition of phArrq is a. (p^^q)vv(p^^q) b. (p=>q)^^(q=>p) c. (p^^q)vv(p=>q) d. (p^^q)(pvvq)

The logically equivalent proposition of phArrq is (p^^q)vv(p^^q) b. (p=>q)^^(p=>q) c. (p^^q)vv(p=>q) d. (p^^q)(pvvq)

Show that [ not p vv not q] vv p is a tautology.

Prove that [(p^^~q)vv(q^^~p)]^^(pvvq)=(pvvq)^^(~qvv~p)^^(pvvq)

~(pvvq)vv(~p^^q) is equivalent to

~(pvvq)vv(~p^^q) is equivalent to