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Proof step 12 from proof mtd-cons-above-fillcount-empty (previous step)

# buf = PEB_SIZE, PEB_SIZE = n0 + m, pebs[n] = mkpeb(buf, m, false), n < # pebs, n1 < m1, is-aligned(m, EB_PAGE_SIZE), m ≤ PEB_SIZE, m ≤ m0, m0 + m1 ≤ PEB_SIZE, ∀ n2. n2 < n0 → buf[m + n2] = empty ⊦ buf[m0 + n1] = empty

User applied rule all left

And obtained the following new goals

Proof step 13:
# buf = PEB_SIZE, PEB_SIZE = n0 + m, pebs[n] = mkpeb(buf, m, false), n1 < m1, n < # pebs, is-aligned(m, EB_PAGE_SIZE), m0 + m1 ≤ PEB_SIZE, m ≤ m0, m ≤ PEB_SIZE, m0 + n1 - m < n0 → buf[m + (m0 + n1 - m)] = empty ⊦ buf[m0 + n1] = empty

Rule argument

pos: left 10 terms: m0 + n1 - m abort: F discard: T computed: F

Additional information

Used terms: m0 + n1 - m