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Proof step 2 from proof lf-valid-wl-free-emptyvidhdr (previous step)

wlar[n] = wl-entry(m, free), n < # apebs, n < # wlar, ubi-io-inv(apebs), valid-wl(apebs, wlar) ⊦ apebs[n].vidhdr = empty ∧ apebs[n].echdr.valid? ∧ apebs[n].fill = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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