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Proof step 7 from proof indices-set-erasing (previous step)

wlar[m0] = wl-entry(m1, erasing), m0 < # wlar, indices-erasing(wlar, nx0), ∀ wlar, m, n. indices-erasing(wlar, nx0) → indices-erasing(wlar[m, wl-entry(n, erasing)], nx0) ⊦ wlar[m, wl-entry(n, erasing)][m0].status = erasing ∧ indices-erasing(wlar[m, wl-entry(n, erasing)], nx0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
wlar[m0] = wl-entry(m1, erasing), wlar[m, wl-entry(n, erasing)][m0] = wl-entry(m2, stat), m0 < # wlar, indices-erasing(wlar, nx0), ∀ wlar, m, n. indices-erasing(wlar, nx0) → indices-erasing(wlar[m, wl-entry(n, erasing)], nx0) ⊦ stat = erasing ∧ indices-erasing(wlar[m, wl-entry(n, erasing)], nx0)

Rule argument

spec: wlentry name: elim-02 seq: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat) subst: [wle, m, stat] → [wlar[m, wl-entry(n, erasing)][m0], m2, stat]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[m, wl-entry(n, erasing)][m0], m2, stat