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Proof step 3 from proof in-index-in-union-1 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ adr ∈ union(asar, n, m) ↔ (∃ n0. n ≤ n0 ∧ n0 < n + m ∧ adr ∈ asar[n0])
Proof step 5:
adr ∈ asar[n], n ≤ m ⊦ # asar[n] = 0, asar[n] = ∅, adr ∈ asar[m], ∃ n0. 0 ≤ n0 ∧ n0 < 0 + m ∧ adr ∈ asar[n0]

Rule argument

spec: <toplevel> name: in-union-ext seq: ⊦ adr ∈ union(asar, n, m) ↔ (∃ n0. n ≤ n0 ∧ n0 < n + m ∧ adr ∈ asar[n0]) subst: [m, n, asar, adr] → [m, 0, asar, adr] flags: <all paths>

Additional information

Used terms: m, 0, asar, adr