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Proof step 14 from proof valid-eraseqw-use (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
wlar[n] = wl-entry(m0, stat), n < # wlar, n ∈ q, ∀ n. n ∈ q ↔ n < # wlar ∧ wlar[n].status = erasing ⊦ stat = erasing

Rule argument

Additional information

Used simplifier rules: ⊦ used ≠ erasing; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ wl-entry(m, stat) = wl-entry(m0, stat0) ↔ m = m0 ∧ stat = stat0; (ws)% ⊦ m < # wlar → wlar[m, wle][m] = wle; (s)