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Proof step 10 from proof in-free-h (previous step)

n ∈ free-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = free ∧ wlar[m1].ec = n), wlar[m] = wl-entry(m2, free), m < # wlar, n = m2 ∨ n ∈ free-ecs(wlar, m) ⊦ #(free-ecs(wlar, m) ++ m2) = 0, ∃ m0. ¬ m < m0 ∧ wlar[m0].status = free ∧ wlar[m0].ec = n

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

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