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Proof step 12 from proof inv-write (previous step)

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

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 13:
pebs[m3] = mkpeb(buf0, m1, false), err0 = ESUCCESS, m3 < # pebs, is-aligned(n, EB_PAGE_SIZE), is-aligned(m2, EB_PAGE_SIZE), mtd-cons(pebs), m2 + n ≤ # pebs[m3].data, n2 + n ≤ # buf, m2 + n ≤ PEB_SIZE, m1 ≤ m2, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪pebs := pebs[m3, pebs[m3].data:= copy(buf, n2, pebs[m3].data, m2, n)] [m3, pebs[m3, pebs[m3].data:= copy(buf, n2, pebs[m3].data, m2, n)][m3].fill:= (m2 + n) ]⟫ mtd-cons(pebs)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ⌊ESUCCESS⌋ ; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ mkpeb(buf, m, boolvar).data = buf; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%n < # pebs ⊦ pebs[n] = mkpeb(buf, m, boolvar) ∧ mtd-cons(pebs) → # buf = PEB_SIZE; (s)% ⊦ ¬ n < n; (s)