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Proof step 8 from proof union-subset (previous step)

union(asar, n, m0) ⊆ as ↔ (∀ n0. n ≤ n0 ∧ n0 < n + m0 → asar[n0] ⊆ as), ∀ n0. n ≤ n0 ∧ ¬ n + m0 < n0 → asar[n0] ⊆ as ⊦ asar[n + m0] ∪ union(asar, n, m0) ⊆ as

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ as ∪ as0 ⊆ as1 ↔ as ⊆ as1 ∧ as0 ⊆ as1
Proof step 10:
union(asar, n, m0) ⊆ as ↔ (∀ n0. n ≤ n0 ∧ n0 < n + m0 → asar[n0] ⊆ as), ∀ n0. n ≤ n0 ∧ ¬ n + m0 < n0 → asar[n0] ⊆ as ⊦ asar[n + m0] ⊆ as ∧ union(asar, n, m0) ⊆ as

Rule argument

spec: set-union, inst: address-list-set name: unsub-union-01 seq: ⊦ as ∪ as0 ⊆ as1 ↔ as ⊆ as1 ∧ as0 ⊆ as1 subst: [as1, as0, as] → [as, union(asar, n, m0), asar[n + m0]] flags: <all paths>

Additional information

Used terms: as, union(asar, n, m0), asar[n + m0]