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Proof step 6 from proof in-union-ext (previous step)

adr ∈ union(asar, n, m0) ↔ (∃ n0. n ≤ n0 ∧ n0 < n + m0 ∧ adr ∈ asar[n0]), adr ∈ asar[n + m0] ∪ union(asar, n, m0) ⊦ #(asar[n + m0] ∪ union(asar, n, m0)) = 0, asar[n + m0] ∪ union(asar, n, m0) = ∅, ∃ n0. n ≤ n0 ∧ ¬ n + m0 < n0 ∧ adr ∈ asar[n0]

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

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