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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ks = ks0 ↔ (∀ key. key ∈ ks ↔ key ∈ ks0)
Proof step 5:
nused < # zbrar, n < # zbrar, sorted(zbrar, 0, nused + 1), n ≤ nused ⊦ ∀ key. key ∈ keys(shl(zbrar, n, nused + 1), 0, nused) ↔ key ∈ keys(zbrar, 0, nused + 1) -- zbrar[n].key

Rule argument

spec: set-basic, inst: node-index name: Extension seq: ⊦ ks = ks0 ↔ (∀ key. key ∈ ks ↔ key ∈ ks0) subst: [ks0, ks] → [keys(zbrar, 0, nused + 1) -- zbrar[n].key, keys(shl(zbrar, n, nused + 1), 0, nused)] flags: <all paths>

Additional information

Used terms: keys(zbrar, 0, nused + 1) -- zbrar[n].key, keys(shl(zbrar, n, nused + 1), 0, nused)