“…If V (G − {a, b, c}) = {s, t, u, v, w} then we may suppose that stu, stv, uvw, abc ∈ G. Since G is K − 4 -free it does not contain suv or tuv. Moreover it contains at most 3 edges from {u, v, w} (2) ×{a, b, c} and at most 5 edges from {s, t, u, v, w}× {a, b, c} (2) . Since G is F 6 -free it contains no edges from {s, t} × {w} × {a, b, c}.…”