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Proof step 3 from proof valid-idx-remlast (previous step)

kar[m] = key-index(key, n), m < # kar, ∀ m0. ¬ m < m0 → kar[m0].idx < # ar ∧ ar[kar[m0].idx].idx = m0 ⊦ valid-idx(kar[0, key-index(key, n)], ar[n, ar[n].idx:= 0], m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ valid-idx(kar, ar, n) ↔ n ≤ # kar ∧ (∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m)
Proof step 5:
kar[m] = key-index(key, n), m < # kar, ∀ m0. ¬ m < m0 → kar[m0].idx < # ar ∧ ar[kar[m0].idx].idx = m0 ⊦ m ≤ # kar[0, key-index(key, n)] ∧ (∀ m0. m0 < m → kar[0, key-index(key, n)][m0].idx < # ar[n, ar[n].idx:= 0] ∧ ar[n, ar[n].idx:= 0][kar[0, key-index(key, n)][m0].idx].idx = m0)

Rule argument

spec: <toplevel> name: valid-idx seq: ⊦ valid-idx(kar, ar, n) ↔ n ≤ # kar ∧ (∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m) subst: [n, ar, kar] → [m, ar[n, ar[n].idx:= 0], kar[0, key-index(key, n)]] flags: <all paths>

Additional information

Used terms: m, ar[n, ar[n].idx:= 0], kar[0, key-index(key, n)]