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Proof step 10 from proof valid-idx-xhg (previous step)

ar[kar[m].idx].idx = m, ar[n0, ar[n0].idx:= m0][n1, ar[n0, ar0[n0].idx:= m0][n1].idx:= m1][m2] = elemindex(a, m3), kar[m0] = key-index(key1, n1), kar[m1] = key-index(key0, n0), kar[m0, key-index(key0, n0)][m1, key-index(key1, n1)][m] = key-index(key, m2), kar[m].idx < # ar, m < # kar, m < n, m0 < # kar, m1 < # kar, m0 < n, m1 < n, n ≤ # kar, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m2 < # ar ∧ m3 = m

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
ar[m4].idx = m, kar[m0, key-index(key0, n0)][m1, key-index(key1, n1)][m] = key-index(key, m2), kar[m1] = key-index(key0, n0), kar[m0] = key-index(key1, n1), ar[n0, ar[n0].idx:= m0][n1, ar[n0, ar0[n0].idx:= m0][n1].idx:= m1][m2] = elemindex(a, m3), kar[m] = key-index(key2, m4), m1 < n, m0 < n, m1 < # kar, m0 < # kar, m < n, m < # kar, m4 < # ar, n ≤ # kar, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m2 < # ar ∧ m3 = m

Rule argument

spec: key-index name: elim seq: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m) subst: [ki, key, m] → [kar[m], key2, m4]

Additional information

Used simplifier rules: ⊦ key = ki.key ∧ m = ki.idx ↔ ki = key-index(key, m); (elim) Used terms: kar[m], key2, m4