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Proof step 12 from proof valid-wl-next-bad (previous step)

# apebs = # wlar, apebs[n] = mkpeb(aehdr0, avhdr0, buf0, m3, boolvar), apebs[m] = mkpeb(aehdr, avhdr, buf, m2, true), wlar[n] = wl-entry(m4, stat), m < # wlar, n < # wlar, n ≤ m, ∀ n. n < m → (wlar[n].status = erroneous ↔ apebs[n].bad) ∧ (wlar[n].status = free → apebs[n].vidhdr = empty) ∧ (wlar[n].status ≠ erroneous ∧ wlar[n].status ≠ erasing → apebs[n].echdr = aechdr(wlar[n].ec)) ∧ (wlar[n].status = used → (∃ v, m1. valid-peb(v, m1, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m1, apebs, m).max)) ⊦ (wlar[m, wl-entry(m0, erroneous)][n].status = erroneous ↔ boolvar) ∧ (wlar[m, wl-entry(m0, erroneous)][n].status = free → avhdr0 = empty) ∧ (wlar[m, wl-entry(m0, erroneous)][n].status ≠ erroneous ∧ stat ≠ erasing → aehdr0 = aechdr(m4)) ∧ ( wlar[m, wl-entry(m0, erroneous)][n].status = used → (∃ v, m1. valid-peb(v, m1, mkpeb(aehdr0, avhdr0, buf0, m3, stat = erroneous)) ∧ avhdr0.sqn = valid-sqns(v, m1, apebs, m).max))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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