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Proof step 10 from proof valid-eraseq-map-indices-nostatchange (previous step)

n ∈ q ↔ n < # wlar ∧ wlar[n].status = erasing, wlar.mapIndices(nx, f)[n] = wl-entry(m, stat), valid-indices(wlar, nx), ∀ wle. f(wle).status = wle.status ⊦ n ∈ q ↔ n < # wlar.mapIndices(nx, f) ∧ stat = erasing, dups(nx)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
n ∈ q ↔ n < # wlar ∧ stat0 = erasing, wlar.mapIndices(nx, f)[n] = wl-entry(m, stat), wlar[n] = wl-entry(m0, stat0), valid-indices(wlar, nx), ∀ wle. f(wle).status = wle.status ⊦ n ∈ q ↔ n < # wlar.mapIndices(nx, f) ∧ stat = erasing, dups(nx)

Rule argument

spec: wlentry name: elim-02 seq: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat) subst: [wle, m, stat] → [wlar[n], m0, stat0]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[n], m0, stat0