Let $\rho$ be the rank function of a matroid $M = \struct{S, \mathscr I}$ on $S$ ==== $\rho$ satisfies $(\text R 1)$ ==== $\rho$ satisfies $(\text R 1)$ follows immediately from Rank of Empty Set is Zero. {{qed|lemma}} ==== $\rho$ satisfies $(\text R 2)$ ==== Let: :$X \subseteq S$ :$y \in S$ From Rank Function is Increasing: :$\map \rho X \le \map \rho {X \cup \set y}$ Let $Y$ be a maximal independent subset of $X \cup \set y$. From Leigh.Samphier/Sandbox/Cardinality of Maximal Independent Subset Equals Rank of Set :$\card Y = \map \rho {X \cup \set y}$ By matroid axiom $(\text I 2)$: :$Y \setminus \set y \in \mathscr I$ We now consider two cases. ===== Case 1: $y \in Y$ ===== We have: {{begin-eqn}} {{eqn | l = \map \rho X       | o = \ge        | r = \card{Y \setminus \set y}       | c = {{Defof|Rank Function (Matroid)|Rank Function}} }} {{eqn | r = \card Y - \card{Y \cap \set y}       | c = Cardinality of Set Difference }} {{eqn | r = \card Y - \card{\set y}       | c = Singleton of Element is Subset }} {{eqn | r = \card Y - 1       | c = Cardinality of Singleton }} {{eqn |ll = \leadsto       | l = \map \rho X + 1        | o = \ge       | r = \card Y       | c = Add one to both sides }} {{end-eqn}} {{qed|lemma}} ===== Case 2: $y \notin Y$ ===== We have: {{begin-eqn}} {{eqn | l = \map \rho X + 1       | o = >        | r = \map \rho X }} {{eqn | o = \ge        | r = \card{Y \setminus \set y}       | c = {{Defof|Rank Function (Matroid)|Rank Function}} }} {{eqn | r = \card Y - \card{Y \cap \set y}       | c = Cardinality of Set Difference }} {{eqn | r = \card Y - \O       | c = Intersection With Singleton is Disjoint if Not Element }} {{eqn | r = \card Y - 0       | c = Cardinality of Empty Set }} {{eqn | r = \card Y }} {{end-eqn}} {{qed|lemma}} In either case: :$\map \rho X + 1 \ge \card Y = \map \rho {X \cup \set y}$ {{qed|lemma}} ==== $\rho$ satisfies $(\text R 3)$ ==== Let: :$X \subseteq S$ :$y, z \in S$ ===== Case 1: $y, z \notin X$ ===== Let $A = X \cup \set{y, z}$. From Rank Function is Increasing: :$\map \rho A \ge \map \rho X$ We now prove the contrapositive statement: :$\map \rho A > \map \rho X \implies$ either $\map \rho {X \cup \set y} > \map \rho X$ or $\map \rho {X \cup \set z} > \map \rho X$ Let $\map \rho A > \map \rho X$. Let $Y$ be a maximal independent subset of $X$. Let $Z$ be a maximal independent subset of $A$. From Leigh.Samphier/Sandbox/Cardinality of Maximal Independent Subset Equals Rank of Set: :$\card Y = \map \rho X < \map \rho A = \card Z$ From Independent Set can be Augmented by Larger Independent Set there exists non-empty set $Z' \subseteq Z \setminus Y$: :$Y \cup Z' \in \mathscr I$ :$\card{Y \cup Z'} = \card Z$ {{AimForCont}}: :$y, z \notin Z'$ We have: {{begin-eqn}} {{eqn | l = Z'       | r = Z' \cap Z }} {{eqn | o = \subseteq       | r = Z' \cap A }} {{eqn | r = Z' \cap \paren{X \cup \set {y, z} } }} {{eqn | r = \paren{Z' \cap  X} \cup \paren{Z' \cap \set {y, z} } }} {{eqn | r = \paren{Z' \cap X} \cup \O }} {{eqn | r = \paren{Z' \cap X} }} {{eqn | o = \subseteq       | r = X }} {{eqn | ll = \leadsto       | l = Y \cup Z'       | o = \subseteq       | r = X }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \card{Y \cup Z'}       | r = \card Z }} {{eqn | r = \map \rho A }} {{eqn | o = >        | r = \map \rho X }} {{eqn | o = \ge        | r = \card{Y \cup Z'} }} {{end-eqn}} This is a contradiction. Hence: :either $y \in Z'$ or $z \in Z'$ {{WLOG}}, assume $y \in Z'$. From matroid axiom $(\text I 2)$: :$Y \cup \set y \in \mathscr I$ We have: {{begin-eqn}} {{eqn | l = \map \rho {X \cup \set y}       | o = \ge       | r = \card{Y \cup \set y}       | c = {{Defof|Rank Function (Matroid)}} }} {{eqn | r = \card Y + \card{\set y}       | c = Corollary of Cardinality of Set Union }} {{eqn | r = \card Y + 1       | c = Cardinality of Singleton }} {{eqn | o = >       | r = \card Y }} {{eqn | r = \map \rho X       | c = {{Defof|Rank Function (Matroid)}} }} {{end-eqn}} This proves the contrapositive statement. {{qed|lemma}} ===== Case 2: Either $y \in X$ or $z \in X$ ===== {{WLOG}}, assume $y \in X$. Let: :$\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X$. We have: :$X \cup \set{y, z} = X \cup \set z$ Hence: :$\map \rho {X \cup \set{y, z}} = \map \rho {X \cup \set z} = \map \rho X$ {{qed|lemma}} In either case: :$\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set{y, z}} = \map \rho X$