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Proof step 6 from proof union-eq-rem-subseteq (previous step)

union(asar, 0, # asar) = as -- adr, n < # asar, adr0 ∈ asar[n] ⊦ # asar[n] = 0, asar[n] = ∅, adr0 ∈ as

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ as = as0 ↔ (∀ adr. adr ∈ as ↔ adr ∈ as0)
Proof step 8:
n < # asar, adr0 ∈ asar[n], ∀ adr0. adr0 ∈ union(asar, 0, # asar) ↔ adr0 ∈ as -- adr ⊦ # asar[n] = 0, asar[n] = ∅, adr0 ∈ as

Rule argument

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

Additional information

Used terms: as -- adr, union(asar, 0, # asar)