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Proof step 4 from proof mtd-cons-extract (previous step)

pebs[n] = mkpeb(buf, m, boolvar), n < # pebs, ∀ n0. n0 < # pebs → # pebs[n0].data = PEB_SIZE ∧ ( ¬ pebs[n0].bad → pebs[n0].fill ≤ PEB_SIZE ∧ is-aligned(pebs[n0].fill, EB_PAGE_SIZE) ∧ range=(pebs[n0].data, pebs[n0].fill, emptypeb, pebs[n0].fill, PEB_SIZE - pebs[n0].fill)) ⊦ # buf = PEB_SIZE ∧ (¬ boolvar → m ≤ PEB_SIZE ∧ is-aligned(m, EB_PAGE_SIZE) ∧ range=(buf, m, emptypeb, m, PEB_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(buf, m, boolvar).data = buf; (s)% ⊦ mkpeb(buf, m, boolvar).bad ↔ boolvar; (s)% ⊦ mkpeb(buf, m, boolvar).fill = m; (s) Used terms: n