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Proof step 18 from proof keys-shift-left (previous step)

zbrar[n0].key = zbrar[n].key, n0 + 1 < # zbrar, n0 < nused, n < # zbrar, nused < # zbrar, n0 < # zbrar, n0 < n, sorted(zbrar, 0, nused + 1), n ≤ nused ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
n + m ≤ # zbrar ⊦ sorted(zbrar, n, m) ↔ (∀ m0, m1. m0 < m1 ∧ m1 < m → zbrar[n + m0].key < zbrar[n + m1].key)
Proof step 20:
zbrar[n0].key = zbrar[n].key, n0 < n, n0 < # zbrar, nused < # zbrar, n < # zbrar, n0 < nused, n0 + 1 < # zbrar, n ≤ nused, 0 + nused + 1 ≤ # zbrar, ∀ m0, m1. m0 < m1 ∧ m1 < nused + 1 → zbrar[0 + m0].key < zbrar[0 + m1].key ⊦

Rule argument

spec: zbranch-array-sorted name: sorted-all seq: n + m ≤ # zbrar ⊦ sorted(zbrar, n, m) ↔ (∀ m0, m1. m0 < m1 ∧ m1 < m → zbrar[n + m0].key < zbrar[n + m1].key) subst: [m, n, zbrar] → [nused + 1, 0, zbrar] flags: <all paths>

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: nused + 1, 0, zbrar