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Proof step 9 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, stat), stat = free, m < # wlar, n ∈ (stat = free ⊃ free-ecs(wlar, m) ++ m2;free-ecs(wlar, m)) ⊦ # (stat = free ⊃ free-ecs(wlar, m) ++ m2;free-ecs(wlar, m)) = 0, free-ecs(wlar, m) ++ m2 = ∅, ∃ m0. ¬ m < m0 ∧ wlar[m0].status = free ∧ wlar[m0].ec = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
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

Rule argument

Additional information

Used simplifier rules: size-zero: ⊦ # ns ≠ 0 → ns ≠ ∅; (forward) from specification set-basic[nat-set]% ⊦ n ∈ ns ++ n0 ↔ n = n0 ∨ n ∈ ns; (s)% ⊦ ∅ ≠ ns ++ n; (s)