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Proof step 2 from proof bflipset-subarray-erase-old (previous step)

n0 < n, n < # wlar, bflipset-cons(bflipset, subarray(wlar, 0, n)) ⊦ bflipset-cons(bflipset -- n0, subarray(wlar[n, wl-entry(m0, used)][n0, wl-entry(m1, erasing)], 0, n + 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 < # wlar, n0 < n, n < # wlar, n0 + 1 < # wlar, n1 ∈ bflipset, ∀ n1. n1 ∈ bflipset → n1 < n ∧ wlar[n1].status = used ⊦ # bflipset = 0, bflipset = ∅, n1 = n0, ¬ n < n1 ∧ wlar[n, wl-entry(m0, used)][n1].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 +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%ne-01: ⊦ n ∈ ns → ns ≠ ∅; (forward) from specification set-basic[nat-set]% ⊦ m ≠ n → wlar[n, wle][m] = wlar[m]; (s)% ⊦ 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)% ⊦ n ∈ ns -- n0 ↔ n ≠ n0 ∧ n ∈ ns; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ # subarray(wlar, m, n) = n; (s)