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Proof step 26 from proof indices-erasing-eraseq (previous step)

m ∈ q1 ↔ m < # wlar1 ∧ wlar1[m].status = erasing, wlar1[m] = wl-entry(m1, stat), wlar0[m] = wl-entry(m0, erasing), m < # wlar0, indices-erasing(wlar1, nx0), indices-erasing(wlar0, nx0), eq-entry(m, lref) ∈ q0, eq-entry(m, lref) ∈ q1 ⊦ m < # wlar1 ∧ stat = erasing

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
wlar1[m] = wl-entry(m1, stat), wlar0[m] = wl-entry(m0, erasing), m < # wlar0, indices-erasing(wlar1, nx0), indices-erasing(wlar0, nx0), eq-entry(m, lref) ∈ q0, eq-entry(m, lref) ∈ q1 ⊦ m ∈ q1, m < # wlar1 ∧ stat = erasing

Rule argument

Additional information

Used simplifier rules: ⊦ wl-entry(m, stat).status = stat; (s)