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Proof step 5 from proof in-index-in-union-1 (previous step)

adr ∈ asar[n], n ≤ m ⊦ # asar[n] = 0, asar[n] = ∅, adr ∈ asar[m], ∃ n0. 0 ≤ n0 ∧ n0 < 0 + m ∧ adr ∈ asar[n0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
adr ∈ asar[n], n ≤ m ⊦ # asar[n] = 0, asar[n] = ∅, adr ∈ asar[m], ∃ n0. n0 < m ∧ adr ∈ asar[n0]

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ 0 ≤ m; (s)