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Proof step 3 from proof valid-eraseqw-free-in (previous step)

wlar[n] = wl-entry(m, free), n < # wlar, n ∈ q, ∀ n. n ∈ q ↔ n < # wlar ∧ wlar[n].status = erasing ⊦

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 4:
# wlar ∈ q ↔ # wlar < # wlar ∧ wlar[# wlar].status = erasing, wlar[n] = wl-entry(m, free), n < # wlar, n ∈ q, ∀ n. n ∈ q ↔ n < # wlar ∧ wlar[n].status = erasing ⊦

Rule argument

pos: left 4 terms: # wlar

Additional information

Used terms: # wlar