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Proof step 2 from proof valid-eraseqw-use (previous step)

wlar[n] = wl-entry(m0, stat) ⊦ valid-eraseq(wlar[n, wl-entry(m1, used)], q) ↔ valid-eraseq(wlar, q), stat = erasing

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
wlar[n] = wl-entry(m0, stat) ⊦ (∀ n0. n0 ∈ q ↔ n0 < # wlar ∧ wlar[n, wl-entry(m1, used)][n0].status = erasing) ↔ (∀ n. n ∈ q ↔ n < # wlar ∧ wlar[n].status = erasing), stat = erasing

Rule argument

Additional information

Used simplifier rules: valid-eraseqw: ⊦ valid-eraseq(wlar, q) ↔ (∀ n. n ∈ q ↔ n < # wlar ∧ wlar[n].status = erasing);% ⊦ # wlar[n, wle] = # wlar; (s)