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Proof step 44 from proof prepost-wl-free-max-below (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
¬ boolvar ↔ below(free-ecs(wlar, # wlar), n) = ∅, wlar[m0] = wl-entry(m1, free), m0 < # wlar, m1 < n, boolvar → vtblpnum < # wlar ∧ wlar[vtblpnum].status = free ∧ wlar[vtblpnum].ec = below(free-ecs(wlar, # wlar), n).max ⊦ vtblpnum < # wlar ∧ wlar[vtblpnum].status = free ∧ wlar[vtblpnum].ec = below(free-ecs(wlar), n).max

Rule argument

Additional information

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