Implication gate is A->B which becomes A'+B
So, let $f(A,B)=A'+B$
$f(A,0)=A'$ (we get complement )
$f(f(A,0),B)=f(A',B)=A+B$ (we get OR gate)
Thus it is functionally complete.
Let $F(X,Y) =X'+Y$
$F(Y,X)=Y'+X$
$F(F(Y'+X),0)=X'Y$
$F(F(X,Y),X'Y)=XY'+XY'$ Therefore, the above function is implemented with $4$ implication gates.