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Proof step 18 from proof in-used-h (previous step)

wlar[m0].status = used, wlar[m] = wl-entry(m2, stat), m0 < # wlar, m0 < m, m < # wlar, m0 + 1 < # wlar, wlar[m0].ec ∈ used-ecs(wlar, m) ⊦ # used-ecs(wlar, m) = 0, stat = used, used-ecs(wlar, m) = ∅, ∃ m1. ¬ m < m1 ∧ wlar[m1].status = used ∧ wlar[m1].ec = wlar[m0].ec

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
wlar[m] = wl-entry(m2, stat), wlar[m0] = wl-entry(m3, stat0), stat0 = used, m0 + 1 < # wlar, m < # wlar, m0 < m, m0 < # wlar, m3 ∈ used-ecs(wlar, m) ⊦ # used-ecs(wlar, m) = 0, stat = used, used-ecs(wlar, m) = ∅, ∃ m1. ¬ m < m1 ∧ wlar[m1].status = used ∧ wlar[m1].ec = m3

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], m3, stat0]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[m0], m3, stat0