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Proof step 25 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, 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 ⊦ wlar[m].status = free ∧ wlar[m].ec < n ∧ (¬ boolvar ∨ wlar[m].ec > wlar[vtblpnum].ec), ( m + 1 ≤ # wlar ∧ (boolvar → vtblpnum < # wlar ∧ wlar[vtblpnum].status = free ∧ wlar[vtblpnum].ec = below(free-ecs(wlar, m + 1), n).max) ∧ (¬ boolvar ↔ below(free-ecs(wlar, m + 1), n) = ∅)) ∧ # wlar - m + 1 < # wlar - m3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
¬ boolvar ↔ below(free-ecs(wlar, m3), n) = ∅, wlar[m0] = wl-entry(m1, free), m0 < # wlar, m1 < n, m3 < # wlar, boolvar → vtblpnum < # wlar ∧ wlar[vtblpnum].status = free ∧ wlar[vtblpnum].ec = below(free-ecs(wlar, m3), n).max, wlar[m3].status = free → wlar[m3].ec < n → boolvar → vtblpnum < # wlar ⊦ (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) = ∅), wlar[m3].status = free ∧ wlar[m3].ec < n ∧ (¬ boolvar ∨ wlar[vtblpnum].ec < wlar[m3].ec)

Rule argument

Additional information

Used simplifier rules: ⊦ m > n ↔ n < m; (s)% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ n < m → (m - n) + n = m; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n → m ≤ n; (s)