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Proof step 37 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 + n1 ≤ # buf, m2 + n1 ≤ # pebs[m3].data, n2 + n ≤ # buf, m2 + n ≤ PEB_SIZE, m1 ≤ m2, n1 < n ∨ n1 = 0 ∧ n = 0, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪pebs := pebs[m3, pebs[m3].data:= copy(buf, n2, pebs[m3].data, m2, n1)] [m3, pebs[m3, pebs[m3].data:= copy(buf, n2, pebs[m3].data, m2, n1)][m3].fill:= (m2 + n1) ]⟫ mtd-cons(pebs), err0 = ESUCCESS, n1 = 0

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 38:
pebs[m3] = mkpeb(buf0, m1, false), m3 < # pebs[m3, pebs[m3].data:= copy(buf, n2, pebs[m3].data, m2, n1)], 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, m2 + n1 ≤ # pebs[m3].data, n2 + n1 ≤ # buf, ⌊err0⌋ ∨ err0 = ESUCCESS, n1 < n ∨ n1 = 0 ∧ n = 0 ⊦ err0 = ESUCCESS, n1 = 0, mtd-cons (pebs[m3, pebs[m3].data:= copy(buf, n2, pebs[m3].data, m2, n1)] [m3, pebs[m3, pebs[m3].data:= copy(buf, n2, pebs[m3].data, m2, n1)][m3].fill:= (m2 + n1) ])

Rule argument

Additional information

Used simplifier rules: ⊦ ⌊err⌋ → err ≠ ESUCCESS; (s)% ⊦ # pebs[n, peb] = # pebs; (s)