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Proof step 20 from proof in-keyrange-in-ri-data (previous step)

n + 1 < nused, n < # zbrar, n < nused, n + 1 < # zbrar, adr ∈ (ri |> zbrar[n - 1].key) |≤ zbrar[n].key, nused ≤ # zbrar ⊦ n + 1 = nused, n = 0, adr ∈ ri

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
n = m + 1, n + 1 < # zbrar, n < nused, n < # zbrar, n + 1 < nused, adr ∈ (ri |> zbrar[m].key) |≤ zbrar[n].key, nused ≤ # zbrar ⊦ n + 1 = nused, n = 0, adr ∈ ri

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m