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Proof step 2 from proof subseteq-union-one (previous step)

n < m, n < # asar, m ≤ # asar ⊦ asar[n] ⊆ union(asar, 0, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0)

Rule argument

spec: set-basic, inst: address-list-set name: Subset seq: ⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0) subst: [as0, as] → [union(asar, 0, m), asar[n]] flags: <all paths>

Additional information

Used simplifier rules: in-index-in-union: ⊦ adr ∈ asar[n] ∧ n < m → adr ∈ union(asar, 0, m); Used terms: union(asar, 0, m), asar[n]