User:Dfeuer/Progressing Function Lemma/General Result

Theorem
Let $A$ be a class.

Let $\preceq$ be an ordering on $A$.

Let $\bot$ be the $\preceq$-smallest element of $A$.

Let $g$ be a $\preceq$-inflationary mapping whose domain and image lie within $A$ and whose domain includes $\bot$.

Define a relation $R$ on the domain of $g$ by letting $xRy \iff (g(x) \preceq y) \lor (y \preceq x)$.

Then for all $x$ and $y$ in $\operatorname{dom}g$:


 * $(1)\quad x R \bot$
 * $(2)\quad x R y \land y R x \implies x R g(y)$

Proof
$(1)$ follows trivially from the choice of $\bot$ and the definition of $R$.

Suppose that $xRy$ and $yRx$.

Then $((g(x) \preceq y) \lor (y \preceq x)) \land ((g(y) \preceq x) \lor (x \preceq y))$.

By Conjunction of Disjunctions Consequence:


 * $(g(y) \preceq x) \lor (g(x) \preceq y) \lor ((x \preceq y) \land (y \preceq x))$

If $g(y) \preceq x$ then $x R g(y)$.

If $g(x) \preceq y$ then since $g$ is inflationary $y \preceq g(y)$.

Since $\preceq$ is transitive, $g(x) \preceq g(y)$, so $x R g(y)$.

If $(x \preceq y) \land (y \preceq x)$ then since $\preceq$ is antisymmetric, $x = y$.

Thus $g(x) =g(y)$.

Since $\preceq$ is reflexive, $g(x) \preceq g(y)$, so $x R g(y)$.