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Proof step 6 from proof valid-eraseqw-use (previous step)

wlar[n] = wl-entry(m0, stat), wlar[n0] = wl-entry(m, stat0), ∀ n0. n0 ∈ q ↔ n0 < # wlar ∧ wlar[n, wl-entry(m1, used)][n0].status = erasing ⊦ n0 ∈ q ↔ n0 < # wlar ∧ stat0 = erasing, stat = erasing

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 7:
n0 ∈ q ↔ n0 < # wlar ∧ wlar[n, wl-entry(m1, used)][n0].status = erasing, wlar[n0] = wl-entry(m, stat0), wlar[n] = wl-entry(m0, stat), ∀ n0. n0 ∈ q ↔ n0 < # wlar ∧ wlar[n, wl-entry(m1, used)][n0].status = erasing ⊦ n0 ∈ q ↔ n0 < # wlar ∧ stat0 = erasing, stat = erasing

Rule argument

pos: left 3 terms: n0

Additional information

Used terms: n0