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Proof step 21 from proof in-used-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
wlar[m0].status = used, wlar[m0].ec ∈ used-ecs(wlar, m) ↔ (∃ m1. m1 < m ∧ wlar[m1].status = used ∧ wlar[m1].ec = wlar[m0].ec), wlar[m] = wl-entry(m2, stat), m < # wlar, m0 < # wlar, m0 ≤ m ⊦ wlar[m0].ec ∈ (stat = used ⊃ used-ecs(wlar, m) ++ m2;used-ecs(wlar, m))

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)