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Proof step 2 from proof bitflip-cons-set-used (previous step)

bflipset-cons(bflipset, wlar) ⊦ bflipset-cons(bflipset, wlar[m, wl-entry(n, used)])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 ∈ bflipset, ∀ n. n ∈ bflipset → n < # wlar ∧ wlar[n].status = used ⊦ # bflipset = 0, bflipset = ∅, n0 < # wlar ∧ wlar[m, wl-entry(n, used)][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);%size-zero-01: ⊦ ns ≠ ∅ → # ns ≠ 0; (forward) from specification set-basic[nat-set]%ne-01: ⊦ n ∈ ns → ns ≠ ∅; (forward) from specification set-basic[nat-set]% ⊦ # wlar[n, wle] = # wlar; (s)