Foundational Relation Subset

Theorem

Suppose $R$ is a foundational relation on $A$. Then if $B \subseteq A$, then $R$ is a foundational relation on $B$.

Proof

Since Subset Relation is Ordering, the subset relation is transitive. Since $B \subseteq A$, $x \subseteq B \implies x \subseteq A$. Therefore,

$\forall x: ( x \not = \varnothing \land x \subseteq A \implies \exists y \in x: \forall z \in x: \neg z R y ) \implies \forall x: ( x \not = \varnothing \land x \subseteq B \implies \exists y \in x: \forall z \in x: \neg z R y )$

By the definition of a foundational relation, $R$ is a foundational relation on $B$.

See Also

• Foundational Relation
• Axiom of Foundation
• No Membership Loops
• Epsilon is Foundational