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Proof step 3 from proof mtd-cons-above-fillcount-empty (previous step)

pebs[n] = mkpeb(buf, m, false), n < # pebs, m ≤ m0, m0 + m1 ≤ PEB_SIZE, ∀ 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)) ⊦ subarray(buf, m0, m1) = mkempbuf(m1)

User applied rule all left

And obtained the following new goals

Proof step 4:
pebs[n] = mkpeb(buf, m, false), n < # pebs, m0 + m1 ≤ PEB_SIZE, m ≤ m0, n < # pebs → # pebs[n].data = PEB_SIZE ∧ ( ¬ pebs[n].bad → pebs[n].fill ≤ PEB_SIZE ∧ is-aligned(pebs[n].fill, EB_PAGE_SIZE) ∧ range=(pebs[n].data, pebs[n].fill, emptypeb, pebs[n].fill, PEB_SIZE - pebs[n].fill)) ⊦ subarray(buf, m0, m1) = mkempbuf(m1)

Rule argument

pos: left 5 terms: n abort: F discard: T computed: T

Additional information

Used terms: n