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Proof step 3 from proof subseteq-union-diff-one (previous step)

n0 < # asar, n < # asar, disj(asar, 0, # asar) ⊦ asar[n] ⊆ (union(asar, 0, # asar) \ asar[n0]) ∪ as ↔ n ≠ n0 ∨ asar[n0] ⊆ 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:
n < # asar, n0 < # asar, disj(asar, 0, # asar) ⊦ (∀ adr. adr ∈ asar[n] → adr ∈ (union(asar, 0, # asar) \ asar[n0]) ∪ as) ↔ n ≠ n0 ∨ asar[n0] ⊆ as

Rule argument

spec: set-basic, inst: address-list-set name: Subset seq: ⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0) subst: [as0, as] → [(union(asar, 0, # asar) \ asar[n0]) ∪ as, asar[n]] flags: <all paths>

Additional information

Used terms: (union(asar, 0, # asar) \ asar[n0]) ∪ as, asar[n]