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Proof step 9 from proof valid-wl-free (previous step)

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

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: m

Additional information

Used simplifier rules: ⊦ m < # wlar → wlar[m, wle][m] = wle; (s)% ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ free ≠ erroneous; (s)% ⊦ free ≠ erasing; (s)% ⊦ wl-entry(m, stat).ec = m; (s)% ⊦ used ≠ free; (s)%m ≠ n ⊦ wlar[m, wle][n] = wlar[n]; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).echdr = aehdr; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s) Used terms: m