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Proof step 5 from proof in-index-in-union (previous step)

adr ∈ asar[n], n ≤ m0, ∀ n, adr, asar. n < m0 ∧ adr ∈ asar[n] → # asar[n] = 0 ∨ asar[n] = ∅ ∨ adr ∈ union(asar, 0, m0) ⊦ # asar[n] = 0, asar[n] = ∅, adr ∈ asar[m0], adr ∈ union(asar, 0, m0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ n ≠ 0 → (m ≤ n ↔ m = n ∨ m < n)
Proof step 7:
adr ∈ asar[n], ∀ n, adr, asar. n < m0 ∧ adr ∈ asar[n] → # asar[n] = 0 ∨ asar[n] = ∅ ∨ adr ∈ union(asar, 0, m0), n = m0 ∨ n < m0 ⊦ # asar[n] = 0, asar[n] = ∅, m0 = 0, adr ∈ union(asar, 0, m0), adr ∈ asar[m0]

Rule argument

spec: nat name: le-rec seq: ⊦ n ≠ 0 → (m ≤ n ↔ m = n ∨ m < n) subst: [n, m] → [m0, n] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ 0 ↔ m = 0; (s) Used terms: m0, n