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Proof step 2 from proof valid-eraseqp-in0 (previous step)

valid-eraseq(wlar, q0 + eq-entry(n, lref) ' + q1) ⊦ wlar[n].status = erasing

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n0. n0 ∈ q0 + eq-entry(n, lref) ' + q1 ↔ n0 < # wlar ∧ wlar[n0].status = erasing ⊦ 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);