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Proof step 12 from proof bitflip-cons-set (previous step)

wlar[n] = wl-entry(m1, stat0), m ∈ bflipset, n ∈ bflipset → n < # wlar ∧ stat0 = used, ∀ n. n ∈ bflipset → n < # wlar ∧ wlar[m, wl-entry(m0, stat)][n].status = used ⊦ # bflipset = 0, bflipset = ∅, n = m, stat = used

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m

Additional information

Used simplifier rules: ⊦ m < # wlar → wlar[m, wle][m] = wle; (s)% ⊦ wl-entry(m, stat).status = stat; (s) Used terms: m