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Proof step 14 from proof at-range (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
2 + m < nused, m + 1 < nused, zbrar[m].key < key, 2 + m < # zbrar, m + 1 < # zbrar, key ∈ ri, nused ≤ # zbrar ⊦ ((ri |> zbrar[m].key) |≤ zbrar[m + 1].key)[key] = ri[key], zbrar[m + 1].key < key

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)