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

valid-eraseq(wlar, q0 + eq-entry(m, lref) ' + q1) ⊦ valid-eraseq(wlar[m, wl-entry(m0, stat)], q0 + q1) ↔ ¬ m ∈ q0 ∧ ¬ m ∈ q1, stat = erasing

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n. n ∈ q0 + eq-entry(m, lref) ' + q1 ↔ n < # wlar ∧ wlar[n].status = erasing ⊦ (∀ n. n ∈ q0 + q1 ↔ n < # wlar ∧ wlar[m, wl-entry(m0, stat)][n].status = erasing) ↔ ¬ m ∈ q0 ∧ ¬ m ∈ q1, 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)