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Proof step 27 from proof prepost-wl-free-max-below (previous step)

¬ boolvar ↔ below(free-ecs(wlar, m3), n) = ∅, wlar[m0] = wl-entry(m1, free), wlar[m3] = wl-entry(m, stat), m3 < # wlar, m1 < n, m0 < # wlar, stat = free → m < n → boolvar → vtblpnum < # wlar, boolvar → vtblpnum < # wlar ∧ wlar[vtblpnum].status = free ∧ wlar[vtblpnum].ec = below(free-ecs(wlar, m3), n).max ⊦ stat = free ∧ m < n ∧ (¬ boolvar ∨ wlar[vtblpnum].ec < m), (boolvar → vtblpnum < # wlar ∧ wlar[vtblpnum].status = free ∧ wlar[vtblpnum].ec = below(free-ecs(wlar, m3 + 1), n).max) ∧ (¬ boolvar ↔ below(free-ecs(wlar, m3 + 1), n) = ∅)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 28:
¬ boolvar ↔ below(free-ecs(wlar, m3), n) = ∅, wlar[m3] = wl-entry(m, stat), wlar[m0] = wl-entry(m1, free), wlar[vtblpnum] = wl-entry(m2, stat0), m0 < # wlar, m1 < n, m3 < # wlar, boolvar → vtblpnum < # wlar ∧ stat0 = free ∧ m2 = below(free-ecs(wlar, m3), n).max, stat = free → m < n → boolvar → vtblpnum < # wlar ⊦ (boolvar → vtblpnum < # wlar ∧ stat0 = free ∧ m2 = below(free-ecs(wlar, m3 + 1), n).max) ∧ (¬ boolvar ↔ below(free-ecs(wlar, m3 + 1), n) = ∅), stat = free ∧ m < n ∧ (¬ boolvar ∨ m2 < m)

Rule argument

spec: wlentry name: elim-02 seq: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat) subst: [wle, m, stat] → [wlar[vtblpnum], m2, stat0]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[vtblpnum], m2, stat0