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Proof step 10 from proof prepost-wl-free-max-below (previous step)

wlar[m].status = free, ¬ boolvar ↔ below(free-ecs(wlar, m), n) = ∅, boolvar0 ↔ false, wlar[m0] = wl-entry(m1, free), m = m3, m4 = 0, vtblpnum0 = m2, wlar[m].ec < n, m < # wlar, m1 < n, m0 < # wlar, m ≤ # wlar, wlar[m].status = free → m < # wlar ∧ (wlar[m].ec < n → boolvar → m < # wlar ∧ vtblpnum < # wlar), boolvar → vtblpnum < # wlar ∧ wlar[vtblpnum].status = free ∧ wlar[vtblpnum].ec = below(free-ecs(wlar, m), n).max, ¬ boolvar ∨ wlar[m].ec > wlar[vtblpnum].ec ⊦ ⟪m0 := m + 1⟫ ( ( m0 ≤ # wlar ∧ (true → m < # wlar ∧ wlar[m].status = free ∧ wlar[m].ec = below(free-ecs(wlar, m0), n).max) ∧ (¬ true ↔ below(free-ecs(wlar, m0), n) = ∅)) ∧ # wlar - m0 < # wlar - m3)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 11:
wlar[m].status = free, boolvar0 ↔ false, ¬ boolvar ↔ below(free-ecs(wlar, m), n) = ∅, wlar[m0] = wl-entry(m1, free), m = m3, m4 = 0, vtblpnum0 = m2, m0 < # wlar, m1 < n, m < # wlar, wlar[m].ec < n, m ≤ # wlar, boolvar → vtblpnum < # wlar ∧ wlar[vtblpnum].status = free ∧ wlar[vtblpnum].ec = below(free-ecs(wlar, m), n).max, wlar[m].status = free → m < # wlar ∧ (wlar[m].ec < n → boolvar → m < # wlar ∧ vtblpnum < # wlar), ¬ boolvar ∨ wlar[m].ec > wlar[vtblpnum].ec ⊦ ( m + 1 ≤ # wlar ∧ (true → m < # wlar ∧ wlar[m].status = free ∧ wlar[m].ec = below(free-ecs(wlar, m + 1), n).max) ∧ (¬ true ↔ below(free-ecs(wlar, m + 1), n) = ∅)) ∧ # wlar - m + 1 < # wlar - m3

Proof step 10 from proof prepost-wl-free-max-below (previous step)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

Additional information