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Proof step 3 from proof bitflip-cons-set-used (previous step)

n0 ∈ bflipset, ∀ n. n ∈ bflipset → n < # wlar ∧ wlar[n].status = used ⊦ # bflipset = 0, bflipset = ∅, n0 < # wlar ∧ wlar[m, wl-entry(n, used)][n0].status = used

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
wlar[m, wl-entry(n, used)][n0] = wl-entry(m0, stat), n0 ∈ bflipset, ∀ n. n ∈ bflipset → n < # wlar ∧ wlar[n].status = used ⊦ # bflipset = 0, bflipset = ∅, n0 < # wlar ∧ stat = used

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, wl-entry(n, used)][n0], m0, stat]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[m, wl-entry(n, used)][n0], m0, stat