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Proof step 12 from proof valid-idx-remlast (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: elem-index name: elim seq: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m) subst: [ei, a, m] → [ar[m3], a0, m4]

Additional information

Used simplifier rules: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m); (elim) Used terms: ar[m3], a0, m4