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Proof step 3 from proof union-eq-rem-subseteq (previous step)

union(asar, 0, # asar) = as -- adr, n < # asar ⊦ asar[n] ⊆ as

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0)
Proof step 5:
union(asar, 0, # asar) = as -- adr, n < # asar ⊦ ∀ adr. adr ∈ asar[n] → adr ∈ as

Rule argument

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

Additional information

Used terms: as, asar[n]