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Proof step 20 from proof valid-idx-move-last (previous step)

ar[m1].idx:= m0 = elemindex(a, m4), kar[ar[m1].idx] = key-index(key, m1), kar[m2] = key-index(key0, m1), m2 < # kar, m1 < # ar, ar[m1].idx < # kar, m2 < ar[m1].idx, m2 + 1 < # kar, ∀ m. ¬ ar[m1].idx < m → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m2 = m0, m4 = m2

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
ar[m1].idx:= m0 = elemindex(a, m4), kar[m2] = key-index(key0, m1), kar[m3] = key-index(key, m1), ar[m1] = elemindex(a0, m3), m2 + 1 < # kar, m2 < m3, m3 < # kar, m1 < # ar, m2 < # kar, ∀ m. ¬ m3 < m → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m2 = m0, m4 = m2

Rule argument

spec: elem-index name: elim seq: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m) subst: [ei, a, m] → [ar[m1], a0, m3]

Additional information

Used simplifier rules: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m); (elim) Used terms: ar[m1], a0, m3