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Proof step 15 from proof id-idx-write (previous step)

# ar0 = # ar1, ar1[n0] = elemindex(ar0[n0].elem, m0), ar0[n, ar0[n].idx:= m][n0] = elemindex(a0, m1), n0 < # ar1 ⊦ ar0[n0].elem = a0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
# ar0 = # ar1, ar0[n, ar0[n].idx:= m][n0] = elemindex(a0, m1), ar1[n0] = elemindex(a, m0), ar0[n0] = elemindex(a, m2), n0 < # ar1 ⊦ 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[n0], a, m2]

Additional information

Used simplifier rules: ⊦ a = ei.elem ∧ m = ei.idx ↔ ei = elemindex(a, m); (elim) Used terms: ar0[n0], a, m2