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Proof step 3 from proof in-keyrange-in-ri (previous step)

n < nused, n < # zbrar, key ∈ key-range(zbrar, n, nused, ri), nused ≤ # zbrar ⊦ key ∈ ri

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n < nused ∧ nused ≤ # zbrar → key-range(zbrar, n, nused, ri) = (n = 0 ⊃ (nused = 1 ⊃ ri;ri |≤ zbrar[0].key);(n + 1 = nused ⊃ ri |> zbrar[n - 1].key;(ri |> zbrar[n - 1].key) |≤ zbrar[n].key))
Proof step 5:
n < # zbrar, n < nused, key ∈ (n = 0 ⊃ (nused = 1 ⊃ ri;ri |≤ zbrar[0].key);(n + 1 = nused ⊃ ri |> zbrar[n - 1].key;(ri |> zbrar[n - 1].key) |≤ zbrar[n].key)), nused ≤ # zbrar ⊦ key ∈ ri

Rule argument

spec: <toplevel> name: range seq: ⊦ n < nused ∧ nused ≤ # zbrar → key-range(zbrar, n, nused, ri) = (n = 0 ⊃ (nused = 1 ⊃ ri;ri |≤ zbrar[0].key) ;(n + 1 = nused ⊃ ri |> zbrar[n - 1].key ;(ri |> zbrar[n - 1].key) |≤ zbrar[n].key ) ) subst: [ri, nused, n, zbrar] → [ri, nused, n, zbrar] flags: <all paths>

Additional information

Used terms: ri, nused, n, zbrar