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Proof step 15 from proof valid-idx-remlast (previous step)

kar[ar[n].idx] = key-index(key, n), kar[0, key-index(key, n)][m4] = key-index(key0, m1), ar[m3] = elemindex(a0, m4), kar[m4] = key-index(key1, m3), ar[n, ar[n].idx:= 0][m1] = elemindex(a, m2), n < # ar, ar[n].idx < # kar, m4 < # kar, m4 < ar[n].idx, m4 + 1 < # kar, m3 < # ar, ∀ m. ¬ ar[n].idx < m → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m1 < # ar ∧ m2 = m4

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
ar[n, ar[n].idx:= 0][m1] = elemindex(a, m2), kar[m4] = key-index(key1, m3), ar[m3] = elemindex(a0, m4), kar[0, key-index(key, n)][m4] = key-index(key0, m1), kar[m0] = key-index(key, n), ar[n] = elemindex(a1, m0), m3 < # ar, m4 + 1 < # kar, m4 < m0, m4 < # kar, m0 < # kar, n < # ar, ∀ m. ¬ m0 < m → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m1 < # ar ∧ m2 = m4

Rule argument

spec: elem-index name: elim seq: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m) subst: [ei, a, m] → [ar[n], a1, m0]

Additional information

Used simplifier rules: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m); (elim) Used terms: ar[n], a1, m0