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Proof step 2 from proof bflipset-subarray-next (previous step)

n < # wlar, bflipset-cons(bflipset, subarray(wlar, 0, n)) ⊦ bflipset-cons(bflipset, subarray(wlar[n, wle], 0, n + 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # wlar, n0 ∈ bflipset, ∀ n0. n0 ∈ bflipset → n0 < n ∧ wlar[n0].status = used ⊦ # bflipset = 0, bflipset = ∅, ¬ n < n0 ∧ wlar[n, wle][n0].status = used

Rule argument

Additional information

Used simplifier rules: bflipset-cons: ⊦ bflipset-cons(bflipset, wlar) ↔ (∀ n. n ∈ bflipset → n < # wlar ∧ wlar[n].status = used);%ls-at-subarray-wlar: n0 ≤ # wlar, n1 < n0 ⊦ subarray(wlar, 0, n0)[n1] = wlar[n1];%size-zero-01: ⊦ ns ≠ ∅ → # ns ≠ 0; (forward) from specification set-basic[nat-set]% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%ne-01: ⊦ n ∈ ns → ns ≠ ∅; (forward) from specification set-basic[nat-set]% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ # wlar[n, wle] = # wlar; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ # subarray(wlar, m, n) = n; (s)