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Proof step 9 from proof bitflip-cons-set (previous step)

wlar[n] = wl-entry(m1, stat0), n ∈ bflipset → n < # wlar ∧ wlar[m, wl-entry(m0, stat)][n].status = used, ∀ n. n ∈ bflipset → n < # wlar ∧ wlar[m, wl-entry(m0, stat)][n].status = used ⊦ stat = used, (n ∈ bflipset → n < # wlar ∧ stat0 = used) ∧ ¬ m ∈ bflipset

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
wlar[n] = wl-entry(m1, stat0), wlar[m, wl-entry(m0, stat)][n] = wl-entry(m2, stat1), n ∈ bflipset → n < # wlar ∧ stat1 = used, ∀ n. n ∈ bflipset → n < # wlar ∧ wlar[m, wl-entry(m0, stat)][n].status = used ⊦ stat = used, (n ∈ bflipset → n < # wlar ∧ stat0 = used) ∧ ¬ m ∈ bflipset

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(m0, stat)][n], m2, stat1]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[m, wl-entry(m0, stat)][n], m2, stat1