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Proof step 16 from proof valid-eraseq-map-indices-nostatchange (previous step)

(n ∈ nx ⊃ f(wlar[n]);wlar[n]) = wl-entry(m, stat), n ∈ q ↔ n < # wlar ∧ stat0 = erasing, wlar[n] = wl-entry(m0, stat0), n < # wlar, valid-indices(wlar, nx), ∀ wle. f(wle).status = wle.status ⊦ n ∈ q ↔ n < # wlar ∧ stat = erasing, dups(nx)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
(n ∈ nx ⊃ f(wl-entry(m0, stat0));wl-entry(m0, stat0)) = wl-entry(m, stat), n ∈ q ↔ stat0 = erasing, wlar[n] = wl-entry(m0, stat0), n < # wlar, valid-indices(wlar, nx), ∀ wle. f(wle).status = wle.status ⊦ stat0 = erasing ↔ stat = erasing, dups(nx)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic