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Proof step 10 from proof valid-eraseqw-use (previous step)

wlar[n] = wl-entry(m0, stat), ∀ n. n ∈ q ↔ n < # wlar ∧ wlar[n].status = erasing ⊦ n0 ∈ q ↔ n0 < # wlar ∧ wlar[n, wl-entry(m1, used)][n0].status = erasing, stat = erasing

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
wlar[n] = wl-entry(m0, stat), wlar[n, wl-entry(m1, used)][n0] = wl-entry(m, stat0), ∀ n. n ∈ q ↔ n < # wlar ∧ wlar[n].status = erasing ⊦ n0 ∈ q ↔ n0 < # wlar ∧ stat0 = erasing, 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[n, wl-entry(m1, used)][n0], m, stat0]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[n, wl-entry(m1, used)][n0], m, stat0