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Proof step 30 from proof prepost-wl-free-max-below (previous step)

¬ boolvar ↔ below(free-ecs(wlar, m3), n) = ∅, wlar[vtblpnum] = wl-entry(m2, stat0), wlar[m0] = wl-entry(m1, free), wlar[m3] = wl-entry(m, stat), m1 < n, m0 < # wlar, m3 < # wlar, stat = free → m < n → boolvar → vtblpnum < # wlar, boolvar → vtblpnum < # wlar ∧ stat0 = free ∧ m2 = below(free-ecs(wlar, m3), n).max ⊦ stat = free ∧ m < n ∧ (¬ boolvar ∨ m2 < m), ( boolvar → vtblpnum < # wlar ∧ stat0 = free ∧ m2 = below((wlar[m3].status = free ⊃ free-ecs(wlar, m3) ++ wlar[m3].ec;free-ecs(wlar, m3)), n).max) ∧ (¬ boolvar ↔ below((wlar[m3].status = free ⊃ free-ecs(wlar, m3) ++ wlar[m3].ec;free-ecs(wlar, m3)), n) = ∅)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
¬ boolvar ↔ below(free-ecs(wlar, m3), n) = ∅, wlar[vtblpnum] = wl-entry(m2, stat0), wlar[m0] = wl-entry(m1, free), wlar[m3] = wl-entry(m, stat), m1 < n, m0 < # wlar, m3 < # wlar, stat = free → m < n → boolvar → vtblpnum < # wlar, boolvar → vtblpnum < # wlar ∧ stat0 = free ∧ m2 = below(free-ecs(wlar, m3), n).max ⊦ stat = free ∧ m < n ∧ (¬ boolvar ∨ m2 < m), (boolvar → vtblpnum < # wlar ∧ stat0 = free ∧ m2 = below((stat = free ⊃ free-ecs(wlar, m3) ++ m;free-ecs(wlar, m3)), n).max) ∧ (¬ boolvar ↔ below((stat = free ⊃ free-ecs(wlar, m3) ++ m;free-ecs(wlar, m3)), n) = ∅)

Rule argument

Additional information

Used simplifier rules: ⊦ wl-entry(m, stat).ec = m; (s)% ⊦ wl-entry(m, stat).status = stat; (s)