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Proof step 3 from proof diff-subseteq-subseteq (previous step)

as1 ⊆ as2 ⊦ as0 \ as1 ⊆ as2 ↔ as0 ⊆ as2

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: set-basic, inst: node-store name: Subset seq: ⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0) subst: [as0, as] → [as2, as0] flags: <all paths>

Additional information

Used terms: as2, as0