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Proof step 35 from proof inv-write (previous step)

pebs[m3] = mkpeb(buf0, m1, false), m3 < # pebs, is-aligned(n1, EB_PAGE_SIZE), is-aligned(n, EB_PAGE_SIZE), is-aligned(m2, EB_PAGE_SIZE), mtd-cons(pebs), n2 + n ≤ # buf, m2 + n ≤ PEB_SIZE, m1 ≤ m2, n1 < n ∨ n1 = 0 ∧ n = 0, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪if n1 ≠ 0 then {pebs[m3].data := copy(buf, n2, pebs[m3].data, m2, n1); pebs[m3].fill := m2 + n1}⟫ mtd-cons(pebs), err0 = ESUCCESS

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 36:
pebs[m3] = mkpeb(buf0, m1, false), m3 < # pebs, is-aligned(m2, EB_PAGE_SIZE), is-aligned(n, EB_PAGE_SIZE), is-aligned(n1, EB_PAGE_SIZE), mtd-cons(pebs), m1 ≤ m2, m2 + n ≤ PEB_SIZE, n2 + n ≤ # buf, ⌊err0⌋ ∨ err0 = ESUCCESS, n1 < n ∨ n1 = 0 ∧ n = 0 ⊦ ⟪pebs[m3].data := copy(buf, n2, pebs[m3].data, m2, n1); pebs[m3].fill := m2 + n1⟫ mtd-cons(pebs), err0 = ESUCCESS, n1 = 0
Proof step 58:
pebs[m3] = mkpeb(buf0, m1, false), n1 = 0, m3 < # pebs, is-aligned(m2, EB_PAGE_SIZE), is-aligned(n, EB_PAGE_SIZE), is-aligned(n1, EB_PAGE_SIZE), mtd-cons(pebs), m1 ≤ m2, m2 + n ≤ PEB_SIZE, n2 + n ≤ # buf, ⌊err0⌋ ∨ err0 = ESUCCESS, n1 < n ∨ n1 = 0 ∧ n = 0 ⊦ ⟪skip⟫ mtd-cons(pebs), err0 = ESUCCESS

Rule argument

pos: right 1

Proof step 35 from proof inv-write (previous step)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

Additional information