"p implies q" is equivalent to "(p and q) or not p", which in turn is equivalent to "(p or not p) and (q or not p)". But "p or not p" is always true, so the implication reduces completely to "not p or q". Negating an implication thus gives "not (not p or q)", which is equivalent to "p and not q".
So
not [(a or not b) implies c] <==> (a or not b) and not c
