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Proof step 3 from proof sorted-shift-left (previous step)

n < # zbrar, nused < # zbrar, sorted(zbrar, 0, nused + 1), n ≤ nused ⊦ sorted(shl(zbrar, n, nused + 1), 0, nused)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
n + m ≤ # zbrar ⊦ sorted(zbrar, n, m) ↔ (∀ m0, m1. m0 < m1 ∧ m1 < m → zbrar[n + m0].key < zbrar[n + m1].key)
Proof step 5:
nused < # zbrar, n < # zbrar, sorted(zbrar, 0, nused + 1), n ≤ nused, 0 + nused ≤ # shl(zbrar, n, nused + 1) ⊦ ∀ m0, m1. m0 < m1 ∧ m1 < nused → shl(zbrar, n, nused + 1)[0 + m0].key < shl(zbrar, n, nused + 1)[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, 0, shl(zbrar, n, nused + 1)] flags: <all paths>

Additional information

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