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Proof step 4 from proof ubi-io-initial-imp-inv (previous step)

apebs[n] = mkpeb(aehdr, avhdr, buf, m, boolvar), n < # apebs, ∀ n. n < # apebs → # apebs[n].data = LEB_SIZE ∧ ( ¬ apebs[n].bad → (apebs[n].echdr = empty → apebs[n].vidhdr = empty) ∧ (apebs[n].echdr ≠ garbage ∧ apebs[n].vidhdr = empty → apebs[n].fill = 0) ∧ apebs[n].fill ≤ LEB_SIZE ∧ is-aligned(apebs[n].fill, EB_PAGE_SIZE) ∧ subarray(apebs[n].data, apebs[n].fill, LEB_SIZE - apebs[n].fill) = mkempbuf(LEB_SIZE - apebs[n].fill)) ∧ (¬ apebs[n].bad → apebs[n].echdr.valid? ∧ apebs[n].vidhdr = empty) ⊦ # buf = LEB_SIZE ∧ ( ¬ boolvar → (aehdr = empty → avhdr = empty) ∧ (aehdr ≠ garbage ∧ avhdr = empty → m = 0) ∧ m ≤ LEB_SIZE ∧ is-aligned(m, EB_PAGE_SIZE) ∧ subarray(buf, m, LEB_SIZE - m) = mkempbuf(LEB_SIZE - m))

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n

Additional information

Used simplifier rules: ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).data = buf; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (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).fill = m; (s) Used terms: n