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Proof step 2 from proof valid-eraseqw-free-in (previous step)

wlar[n] = wl-entry(m, free), n < # wlar, valid-eraseq(wlar, q), n ∈ q ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
wlar[n] = wl-entry(m, free), n < # wlar, n ∈ q, ∀ n. n ∈ q ↔ n < # wlar ∧ wlar[n].status = erasing ⊦

Rule argument

Additional information

Used simplifier rules: valid-eraseqw: ⊦ valid-eraseq(wlar, q) ↔ (∀ n. n ∈ q ↔ n < # wlar ∧ wlar[n].status = erasing);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic