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Proof step 2 from proof in-index-in-union-1 (previous step)

adr ∈ asar[n], n ≤ m ⊦ adr ∈ union(asar, 0, m + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
adr ∈ asar[n], n ≤ m ⊦ # asar[n] = 0, asar[n] = ∅, adr ∈ union(asar, 0, m), adr ∈ asar[m]

Rule argument

Additional information

Used simplifier rules: union-rec: ⊦ union(asar, n, m + 1) = asar[n + m] ∪ union(asar, n, m);%size-zero-01: ⊦ as ≠ ∅ → # as ≠ 0; (forward) from specification set-basic[address-list-set]%ne-01: ⊦ adr ∈ as → as ≠ ∅; (forward) from specification set-basic[address-list-set]%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)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)