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Proof step 19 from proof prepost-wl-get-free-min (previous step)

¬ boolvar ↔ free-ecs(wlar, m) = ∅, boolvar0 ↔ false, m = m0, m1 = 0, m < # wlar, m ≤ # wlar, wlar[m].status = free → boolvar → m < # wlar, boolvar → n = free-ecs(wlar, m).min ⊦ wlar[m].status = free ∧ (¬ boolvar ∨ wlar[m].ec < n), (m + 1 ≤ # wlar ∧ (boolvar → n = free-ecs(wlar, m + 1).min) ∧ (¬ boolvar ↔ free-ecs(wlar, m + 1) = ∅)) ∧ # wlar - m + 1 < # wlar - m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
¬ boolvar ↔ free-ecs(wlar, m0) = ∅, m0 < # wlar, boolvar → n = free-ecs(wlar, m0).min ⊦ (boolvar → n = free-ecs(wlar, m0 + 1).min) ∧ (¬ boolvar ↔ free-ecs(wlar, m0 + 1) = ∅), wlar[m0].status = free ∧ (¬ boolvar ∨ wlar[m0].ec < n)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ ¬ 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)