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Proof step 3 from proof id-idx-trans (previous step)

# ar2 = # ar0, # ar1 = # ar0, n < # ar0, ∀ n. n < # ar0 → ar1[n].elem = ar2[n].elem, ∀ n. n < # ar0 → ar0[n].elem = ar1[n].elem ⊦ ar0[n].elem = ar2[n].elem

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
# ar1 = # ar0, # ar2 = # ar0, ar0[n] = elemindex(a, m), n < # ar0, ∀ n. n < # ar0 → ar0[n].elem = ar1[n].elem, ∀ n. n < # ar0 → ar1[n].elem = ar2[n].elem ⊦ a = ar2[n].elem

Rule argument

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

Additional information

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