[N[Np^^(P hArr q)]=],[A)q^^pquad B)pvqquad C)sim pquad D=Nq]

sim(p hArr q) is (A) sim p^^q(C)(p^^-q)vv(-p^^q)(B)-p vv sim q(D) none of these

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]

Which one of the following is not a contradiction? [sim p^^(p vv sim q)]^^q b.(sim p^^q)vv p c.[(p rarr q)rarr]^^sim p d.(sim q rarr sim p)harr(p rarr q)

let (i) (p vv q)^^(p vv sim q),(ii)(p^^q)^^(q vv sim q),(iii).(p vv q)^^(p vv sim q),(iv)(p vv q)vv(p^^sim q) then which one is tautology

The logically equivalent proposition of p hArr q is (p^^q)vv(p^^q) b.(p rArr q)^^(p rArr q) c.(p^^q)vv(p rArr q) d.(p^^q)o+(p vv q)

Show that ~(p harr q) -= ( p ^^ ~q) vv(~p ^^q) .

Show that ~(p harr q) -= ( p ^^ ~q) vv(~p ^^q) .

Show that ~(p harr q) -= ( p ^^ ~q) vv(~p ^^q) .

((p^^q)vv(pvv~q))^^(~p^^~q)) is equivalent to (A) ~p^^~q (B) p^^~q (C) pvv~q (D) ~pvvq

The statement sim(sim p rarr q) is equivalent to (A)p vv sim q(B)sim p^^q(C)sim p^^sim q(D)sim p vv q