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Proof step 6 from proof indices-erasing-at (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
wlar[n] = wl-entry(m, stat), wlar[m0] = wl-entry(m1, stat0), stat0 = erasing, m0 < # wlar, indices-erasing(wlar, nx0), ∀ wlar, n, m, stat. wlar[n] = wl-entry(m, stat) ∧ indices-erasing(wlar, nx0) ∧ n ∈ nx0 → stat = erasing, n = m0 ∨ n ∈ nx0 ⊦ stat = erasing

Rule argument

spec: wlentry name: elim-02 seq: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat) subst: [wle, m, stat] → [wlar[m0], m1, stat0]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[m0], m1, stat0