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Proof step 3 from proof in-used-h (previous step)

m < # wlar, ∀ wlar, n. m ≤ # wlar → (n ∈ used-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = used ∧ wlar[m1].ec = n)) ⊦ n ∈ (wlar[m].status = used ⊃ used-ecs(wlar, m) ++ wlar[m].ec;used-ecs(wlar, m)) ↔ (∃ m0. ¬ m < m0 ∧ wlar[m0].status = used ∧ wlar[m0].ec = n)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
wlar[m] = wl-entry(m2, stat), m < # wlar, ∀ wlar, n. m ≤ # wlar → (n ∈ used-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = used ∧ wlar[m1].ec = n)) ⊦ n ∈ (stat = used ⊃ used-ecs(wlar, m) ++ m2;used-ecs(wlar, m)) ↔ (∃ m0. ¬ m < m0 ∧ wlar[m0].status = used ∧ wlar[m0].ec = n)

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], m2, stat]

Additional information

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