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Proof step 6 from proof inter-union-empty-one-empty (previous step)

as ∩ union(asar, 0, m) = ∅, n < # asar, n < m, adr ∈ asar[n], adr ∈ as, m ≤ # asar ⊦ #(as ∩ asar[n]) = 0, # asar[n] = 0, # as = 0, asar[n] = ∅, 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 < m, n < # asar, adr ∈ as, adr ∈ asar[n], m ≤ # asar, ∀ adr. adr ∈ as ∩ union(asar, 0, m) ↔ adr ∈ ∅ ⊦ # as = 0, # asar[n] = 0, #(as ∩ asar[n]) = 0, asar[n] = ∅, as = ∅

Rule argument

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

Additional information

Used terms: ∅, as ∩ union(asar, 0, m)