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Proof step 4 from proof id-idx-write-left (previous step)

# ar0 = # ar1, n < # ar1, ∀ n. n < # ar1 → ar1[m, ar1[m].idx:= m0][n].elem = ar0[n].elem ⊦ ar1[n].elem = ar0[n].elem

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
# ar0 = # ar1, ar1[n] = elemindex(a, m1), n < # ar1, ∀ n. n < # ar1 → ar1[m, ar1[m].idx:= m0][n].elem = ar0[n].elem ⊦ a = ar0[n].elem

Rule argument

spec: elem-index name: elim seq: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m) subst: [ei, a, m] → [ar1[n], a, m1]

Additional information

Used simplifier rules: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m); (elim) Used terms: ar1[n], a, m1