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Proof step 13 from proof union-write (previous step)

n < # asar, disj(asar, 0, n + 1), ∀ n0, asar, as. n0 < n ∧ n0 < # asar ∧ disj(asar, 0, n) ∧ n ≤ # asar → union(asar[n0, as], 0, n) = (union(asar, 0, n) \ asar[n0]) ∪ as ⊦ #(union(asar, 0, n) ∩ asar[n]) = 0, union(asar, 0, n) ∩ asar[n] = ∅, as ∪ union(asar, 0, n) = (union(asar, 0, n) \ asar[n]) ∪ as

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ as = as0 ↔ (∀ adr. adr ∈ as ↔ adr ∈ as0)
Proof step 15:
n < # asar, disj(asar, 0, n + 1), ∀ n0, asar, as. n0 < n ∧ n0 < # asar ∧ disj(asar, 0, n) ∧ n ≤ # asar → union(asar[n0, as], 0, n) = (union(asar, 0, n) \ asar[n0]) ∪ as ⊦ #(union(asar, 0, n) ∩ asar[n]) = 0, as ∪ union(asar, 0, n) = (union(asar, 0, n) \ asar[n]) ∪ as, ∀ adr. adr ∈ union(asar, 0, n) ∩ asar[n] ↔ adr ∈ ∅

Rule argument

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

Additional information

Used terms: ∅, union(asar, 0, n) ∩ asar[n]