Kuratowski Theorem from below

Proof. Let us assume that G is planar but not free planar. Then there exists an edge xy not belonging to the graph whom adding to the graph it becomes non-planar. Then in G for an arbitrary cycle C through x, y there exists a pair of screening bridges Bx and By x from y with respect to C, i. e. either Bx and By are not placeable on one side against C or they are connected [i.e. not placeable together] with an alternating [i.e. on one and other side of C] sequence [B1, ..., B2k, k > 0] of non-screening [x from y] bridges. Let us describe the bridge with the sextet [x, a, b, y, c, d], where values of it are either vertices on the cycle C or logical values T (= true) or F (= false) [see fig. 1]: 1) in the place of x(y) stands T if x(y) is a leg [i.e. the touch vertex to C] of the bridge with respect to C, otherwise F ; 2) a(c) is the nearest next leg moving clockwise from x(y) before y(x) if any, otherwise F ; 3) b(d) is the nearest next leg moving anticlockwise from y(x) before x(y) if any, otherwise F ; The screening condition of a bridge [x, a, b, y, c, d] x from y on C is – the values a, b, c, d are not F . Non-screening bridges Bi, [0 < i ≤ 2k] are of the form [x, a, b, y, F, F ] or [x, F, F, y, c, d] in general. There are three simple [k = 0] cases and one non-simple case [k > 0] to be considered: 1) In the first case, for one of bridges, say, Bx both in x and y stand T . K − 5 arises even when By is simple: [T, a, a, T, c, c].