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Proof step 14 from proof in-free-h (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
wlar[m] = wl-entry(m2, free), wlar[m0] = wl-entry(m3, stat), stat = free, m0 + 1 < # wlar, m0 < m, m < # wlar, m0 < # wlar, m3 ∈ free-ecs(wlar, m) ⊦ # free-ecs(wlar, m) = 0, #(free-ecs(wlar, m) ++ m2) = 0, free-ecs(wlar, m) = ∅, ∃ m1. ¬ m < m1 ∧ wlar[m1].status = free ∧ 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, stat]

Additional information

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