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Proof step 3 from proof valid-eraseq-prev-erasing (previous step)

m1 < # wlar, m1 < m, m < # wlar, m1 + 1 < # wlar, ∀ n. n ∈ q ↔ n < m ∧ wlar[n].status = erasing ⊦ n ∈ q + eq-entry(m1, lref) ' ↔ ¬ m < n ∧ wlar[m, wl-entry(m0, used)][m1, wl-entry(m2, erasing)][n].status = erasing

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
wlar[m, wl-entry(m0, used)][m1, wl-entry(m2, erasing)][n] = wl-entry(m3, stat), m1 + 1 < # wlar, m < # wlar, m1 < m, m1 < # wlar, ∀ n. n ∈ q ↔ n < m ∧ wlar[n].status = erasing ⊦ n ∈ q + eq-entry(m1, lref) ' ↔ ¬ m < n ∧ stat = erasing

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, used)][m1, wl-entry(m2, erasing)][n], m3, stat]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[m, wl-entry(m0, used)][m1, wl-entry(m2, erasing)][n], m3, stat