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Proof step 2 from proof bitflip-cons-add (previous step)

bflipset-cons(bflipset, wlar) ⊦ bflipset-cons(bflipset ++ n, wlar) ↔ n < # wlar ∧ wlar[n].status = used

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n. n ∈ bflipset → n < # wlar ∧ wlar[n].status = used ⊦ (∀ n0. n0 = n ∨ n0 ∈ bflipset → n0 < # wlar ∧ wlar[n0].status = used) ↔ n < # wlar ∧ wlar[n].status = used

Rule argument

Additional information

Used simplifier rules: bflipset-cons: ⊦ bflipset-cons(bflipset, wlar) ↔ (∀ n. n ∈ bflipset → n < # wlar ∧ wlar[n].status = used);% ⊦ n ∈ ns ++ n0 ↔ n = n0 ∨ n ∈ ns; (s)