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Proof step 12 from proof id-idx-write-left (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
# ar0 = # ar1, ar1[m, ar1[m].idx:= m0][n] = elemindex(a, m1), ar0[n] = elemindex(a0, m2), n < # ar1, ∀ n. n < # ar1 → ar1[n].elem = ar0[n].elem ⊦ a = a0

Rule argument

spec: elem-index name: elim seq: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m) subst: [ei, a, m] → [ar0[n], a0, m2]

Additional information

Used simplifier rules: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m); (elim) Used terms: ar0[n], a0, m2