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Proof step 16 from proof key-range-replace-branches (previous step)

# zbrar0 = # zbrar1, n + 1 < m, n < m, n < # zbrar1, n + 1 < # zbrar1, m ≤ # zbrar1, ∀ n0. n0 ≤ n ∧ n0 ≤ m → zbrar0[n0].key = zbrar1[n0].key ⊦ n + 1 = m, n = 0, (ri |> zbrar0[n - 1].key) |≤ zbrar0[n].key = (ri |> zbrar1[n - 1].key) |≤ zbrar1[n].key

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
# zbrar0 = # zbrar1, n = m0 + 1, n + 1 < # zbrar1, n < # zbrar1, n < m, n + 1 < m, m ≤ # zbrar1, ∀ n0. n0 ≤ n ∧ n0 ≤ m → zbrar0[n0].key = zbrar1[n0].key ⊦ n + 1 = m, n = 0, (ri |> zbrar0[m0].key) |≤ zbrar0[n].key = (ri |> zbrar1[m0].key) |≤ zbrar1[n].key

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m0]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0