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Proof step 43 from proof prepost-wl-free-max-below (previous step)

¬ boolvar ↔ below(free-ecs(wlar, m), n) = ∅, boolvar0 ↔ false, wlar[m0] = wl-entry(m1, free), m = m3, m4 = 0, vtblpnum0 = m2, m < # wlar, m1 < n, m0 < # wlar, m ≤ # wlar, boolvar → vtblpnum < # wlar ∧ wlar[vtblpnum].status = free ∧ wlar[vtblpnum].ec = below(free-ecs(wlar, m), n).max ⊦ m < # wlar ∧ (wlar[m].status = free → m < # wlar ∧ (wlar[m].ec < n → boolvar → m < # wlar ∧ vtblpnum < # wlar))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → m ≤ n; (s)