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Proof step 10 from proof in-keyrange-in-ri (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 3

Additional information

Used simplifier rules: ⊦ key0 ∈ ri |≤ key1 ↔ key0 ∈ ri ∧ ¬ key1 < key0; (s)% ⊦ key0 ∈ ri |> key1 ↔ key0 ∈ ri ∧ key1 < key0; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic