=== Existence of $z$ and $w$ === First, it must be proven that such $z$ and $w$ exist. By Relation between Two Ordinals, it follows that: :$x < y$ or $y \le x$ The proof shall proceed by cases. ==== Case 1 ==== If $x < y$, then set $z = 0$ and $w = x$. Since $w = x$, then $w < y$. Furthermore, by Ordinal Multiplication by Zero: :$x = \paren {y \times z} + w$ ==== Case 2 ==== If $y \le x$, then set $z = \bigcup \set {v : \paren {y \times v} \le x}$. Set $w$ equal to the unique $u$ such that $\paren {y \times z} + u = x$. By Ordinal Multiplication by One, it follows that $1 \le z$. By Union of Ordinals is Least Upper Bound, it follows that $z$ is the least upper bound of $\set {v : \paren {y \times v} \le x}$. Take any $u \in z$. Then $u \in v$ for some $\paren {y \times v} \le x$ But then, by Membership is Left Compatible with Ordinal Multiplication, it follows that $\paren {y \times u} \le x$. So $z$ is a transitive subset of $\On$ and therefore, it is an ordinal. Since $z \ne 0$, then by the definition of limit ordinal, then $z = u^+$ for some $u$ or $z$ is a limit ordinal. Suppose $z = u^+$. Then $u < z$ and $y \times u \le x$. Moreover, $u < v$ for some $y \times v \le x$. This means that $v \not < z$ by No Natural Number between Number and Successor. By Ordinal Membership is Trichotomy, $z \le v$. Therefore: :$y \times z \le x$ Suppose $z \in K_{II}$. Note that $\forall u \in z: y \times u \le x$ so: {{begin-eqn}} {{eqn | l = y \times z       | r = \bigcup_{u \mathop \in z} \paren {y \times u}       | c = {{Defof|Ordinal Multiplication}} }} {{eqn | o = \le       | r = \bigcup_{u \mathop \in z} x       | c = Indexed Union Subset }} {{eqn | r = x       | c = Union is Idempotent }} {{end-eqn}} In either case, $y \times z \le x$. Moreover, suppose that $y \times z^+ \le x$. Then $z$ is not an upper bound of the set $\set {v : \paren {y \times v} \le x}$, which is a contradiction. So $\paren {y \times z} + w = x$ for some $w \in y$ by Ordinal Subtraction when Possible is Unique. === Uniqueness of $z$ and $w$ === Suppose that $\paren {y \times z_1} + w_1 = x$ for some $w_1 \in y$. Suppose that $\paren {y \times z_2} + w_2 = x$ for some $w_2 \in y$. Then for both $z = z_1$ and $z = z_2$, the following inequality is established: {{begin-eqn}} {{eqn | l = y \times z       | o = \le       | r = x       | c = Ordinal is Less than Sum }} {{eqn | o = <       | r = \left({ y \times z }\right) + y       | c = Membership is Left Compatible with Ordinal Addition }} {{eqn | r = y \times z^+       | c = {{Defof|Ordinal Multiplication}} }} {{end-eqn}} It follows that $y \times z_1 < y \times z_2^+$ and $y \times z_2 < y \times z_1^+$. By Membership is Left Compatible with Ordinal Multiplication, $z_1 < z_2^+$ and $z_2 < z_1^+$. {{begin-eqn}} {{eqn | l = z_1 < z_2^+       | o = \leadsto       | r = z_1 < z_2 \lor z_1 = z_2       | c = {{Defof|Successor Set}} }} {{eqn | o = \leadsto       | r = z_1 \le z_2       | c =  }} {{eqn | l = z_2 < z_1^+       | o = \leadsto       | r = z_2 \le z_1       | c = {{Defof|Successor Set}} }} {{eqn | o = \leadsto       | r = z_1 = z_2       | c = {{Defof|Set Equality|index = 2}} }} {{end-eqn}} Therefore $z$ is unique. Since $z$ is unique, then $w$ is unique by Ordinal Subtraction when Possible is Unique. {{qed}}