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Proof step 19 from proof align-down-app-same (previous step)

# buf0 = m0 * EB_PAGE_SIZE + n0, n0 < EB_PAGE_SIZE, n < ((# buf0 + # buf1) % EB_PAGE_SIZE = 0 ⊃ # buf0 + # buf1;(# buf0 + # buf1) / EB_PAGE_SIZE * EB_PAGE_SIZE), n0 + # buf1 < EB_PAGE_SIZE, (n0 = 0 ⊃ # buf0;m0 * EB_PAGE_SIZE) < ((# buf0 + # buf1) % EB_PAGE_SIZE = 0 ⊃ # buf0 + # buf1;(# buf0 + # buf1) / EB_PAGE_SIZE * EB_PAGE_SIZE), (n0 = 0 ⊃ # buf0;m0 * EB_PAGE_SIZE) ≤ n ⊦

User applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
# buf0 = m0 * EB_PAGE_SIZE + n0, # buf0 + # buf1 = m1 * EB_PAGE_SIZE + n1, (n0 = 0 ⊃ # buf0;m0 * EB_PAGE_SIZE) < (n1 = 0 ⊃ # buf0 + # buf1;m1 * EB_PAGE_SIZE), n0 + # buf1 < EB_PAGE_SIZE, n < (n1 = 0 ⊃ # buf0 + # buf1;m1 * EB_PAGE_SIZE), n0 < EB_PAGE_SIZE, n1 < EB_PAGE_SIZE, (n0 = 0 ⊃ # buf0;m0 * EB_PAGE_SIZE) ≤ n ⊦ EB_PAGE_SIZE = 0

Rule argument

spec: nat-div name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [# buf0 + # buf1, EB_PAGE_SIZE, m, n1]

Additional information

Used simplifier rules: ⊦ EB_PAGE_SIZE ≠ 0; (s)%n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: # buf0 + # buf1, EB_PAGE_SIZE, m, n1