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Proof step 9 from proof no-dups-idx-rec (previous step)

n < # kar, key-index(key0, m2) ∈ keyset(kar, n), ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx, ∀ m, m0. m < # kar ∧ ¬ n < m ∧ m0 < # kar ∧ ¬ n < m0 ∧ m ≠ m0 → kar[n, key-index(key, m2)][m].idx ≠ kar[n, key-index(key, m2)][m0].idx ⊦ # keyset(kar, n) = 0, keyset(kar, n) = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
n ≤ # kar ⊦ ki ∈ keyset(kar, n) ↔ (∃ m. m < n ∧ ki = kar[m])
Proof step 11:
n < # kar, n ≤ # kar, ∀ m, m0. m < # kar ∧ ¬ n < m ∧ m0 < # kar ∧ ¬ n < m0 ∧ m ≠ m0 → kar[n, key-index(key, m2)][m].idx ≠ kar[n, key-index(key, m2)][m0].idx, ∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx, ∃ m. m < n ∧ key-index(key0, m2) = kar[m] ⊦ # keyset(kar, n) = 0, 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(key0, m2)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) Used terms: n, kar, key-index(key0, m2)