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Proof step 22 from proof in-free-h (previous step)

wlar[m0].status = free, wlar[m0].ec ∈ free-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = free ∧ wlar[m1].ec = wlar[m0].ec), wlar[m] = wl-entry(m2, stat), m < # wlar, m0 < # wlar, m0 ≤ m ⊦ wlar[m0].ec ∈ (stat = free ⊃ free-ecs(wlar, m) ++ m2;free-ecs(wlar, m))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 23:
m3 ∈ free-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = free ∧ wlar[m1].ec = m3), wlar[m] = wl-entry(m2, stat), wlar[m0] = wl-entry(m3, stat0), stat0 = free, m0 < # wlar, m < # wlar, m0 ≤ m ⊦ m3 ∈ (stat = free ⊃ free-ecs(wlar, m) ++ m2;free-ecs(wlar, m))

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