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Proof step 4 from proof in-index-in-union (previous step)

n < m0 +1, adr ∈ asar[n], ∀ n, adr, asar. n < m0 ∧ adr ∈ asar[n] → # asar[n] = 0 ∨ asar[n] = ∅ ∨ adr ∈ union(asar, 0, m0) ⊦ # asar[n] = 0, asar[n] = ∅, adr ∈ union(asar, 0, m0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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)

Rule argument

Additional information

Used simplifier rules: union-rec: ⊦ union(asar, n, m + 1) = asar[n + m] ∪ union(asar, n, m);% ⊦ m < n ↔ ¬ n ≤ m; (ss)%unc: ⊦ as ∪ as0 = as0 ∪ as; from specification set-union[address-list-set]%una: ⊦ (as ∪ as0) ∪ as1 = as ∪ as0 ∪ as1; from specification set-union[address-list-set]% ⊦ adr ∈ as ∪ as0 ↔ ¬(¬ adr ∈ as ∧ ¬ adr ∈ as0); (ss)% ⊦ n + 0 = n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m +1 = m + 1; (s)