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Proof step 3 from proof key-index-in-keyset-array-extract (previous step)

ar[m] = elemindex(a, m0), key-index(key, m) ∈ keyset(kar, n), n ≤ # kar, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ # keyset(kar, n) = 0, kar[m0] = key-index(key, m), keyset(kar, n) = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
n ≤ # kar ⊦ ki ∈ keyset(kar, n) ↔ (∃ m. m < n ∧ ki = kar[m])
Proof step 5:
ar[m] = elemindex(a, m0), n ≤ # kar, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m, ∃ m0. m0 < n ∧ key-index(key, m) = kar[m0] ⊦ # keyset(kar, n) = 0, kar[m0] = key-index(key, m), keyset(kar, n) = ∅

Rule argument

spec: <toplevel> name: key-set-extract-index seq: n ≤ # kar ⊦ ki ∈ keyset(kar, n) ↔ (∃ m. m < n ∧ ki = kar[m]) subst: [n, kar, ki] → [n, kar, key-index(key, m)] flags: <all paths>

Additional information

Used terms: n, kar, key-index(key, m)