# Membership is Left Compatible with Ordinal Addition

## Theorem

Let $x$, $y$, and $z$ be ordinals.

Let $<$ denote membership $\in$, since $\in$ is a strict well-ordering on the ordinals.

Then:

$x < y \implies \paren {z + x} < \paren {z + y}$

## Proof

The proof proceeds by transfinite induction on $y$.

In the proof, we shall use $\in$, $\subsetneq$, and $<$ interchangeably.

We are justified in this by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.

### Base Case

 $\displaystyle \neg x$ $=$ $\displaystyle \O$ Definition of Empty Set

The conclusion:

$x < \O \implies \paren {z + x} < \paren {z + \O}$

follows from propositional logic.

### Inductive Case

 $\displaystyle x < y$ $\leadsto$ $\displaystyle \paren {z + x} < \paren {z + y}$ Hypothesis $\displaystyle x < y^+$ $\leadsto$ $\displaystyle x < y \lor x = y$ Definition of Successor Set $\displaystyle \paren {z + y^+} = \paren {z + y}^+$  $\displaystyle$ Definition of Ordinal Addition $\displaystyle \paren {z + y} < \paren {z + y^+}$  $\displaystyle$ Ordinal is Less than Successor $\displaystyle x < y$ $\leadsto$ $\displaystyle \paren {z + x} < \paren {z + y^+}$ Hypothetical Syllogism $\displaystyle x = y$ $\leadsto$ $\displaystyle \paren {z + x} < \paren {z + y^+}$ Substitutivity of Equality $\displaystyle x < y^+$ $\leadsto$ $\displaystyle \paren {z + x} < \paren {z + y^+}$ Proof by Cases

### Limit Case

 $\displaystyle$  $\displaystyle \forall w < y: \paren {x < w \implies \paren {z + x} < \paren {z + w} }$ $\displaystyle$ $\leadsto$ $\displaystyle \paren {\exists w < y: x < w \implies \exists w < y \paren {z + x} < \paren {z + w} }$ Predicate Logic Manipulation $\displaystyle$ $\leadsto$ $\displaystyle \paren {x < y \implies \paren {z + x} < \bigcup_{w \mathop \in y} \paren {z + w} }$ Membership of Indexed Union $\displaystyle$ $\leadsto$ $\displaystyle \paren {x < y \implies \paren {z + x} < \paren {z + y} }$ Definition of Ordinal Addition

$\blacksquare$