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Proof step 3 from proof disj-union-rem (previous step)

n0 < m, n1 < m, n0 < # asar, n1 < # asar, disj(asar, 0, m), m ≤ # asar ⊦ (asar[n0] ∪ as) \ asar[n1] = asar[n0] ∪ (as \ asar[n1]), n0 = n1

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:
n1 < # asar, n0 < # asar, n1 < m, n0 < m, disj(asar, 0, m), m ≤ # asar ⊦ n0 = n1, ∀ adr. adr ∈ (asar[n0] ∪ as) \ asar[n1] ↔ adr ∈ asar[n0] ∪ (as \ asar[n1])

Rule argument

spec: set-basic, inst: address-list-set name: Extension seq: ⊦ as = as0 ↔ (∀ adr. adr ∈ as ↔ adr ∈ as0) subst: [as0, as] → [asar[n0] ∪ (as \ asar[n1]), (asar[n0] ∪ as) \ asar[n1]] flags: <all paths>

Additional information

Used terms: asar[n0] ∪ (as \ asar[n1]), (asar[n0] ∪ as) \ asar[n1]