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Proof step 20 from proof align-down-app-same (previous step)

# 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# buf0 = m0 * EB_PAGE_SIZE + n0, # buf0 + # buf1 = m1 * EB_PAGE_SIZE + n1, (n0 = 0 ⊃ # buf0;m0 * EB_PAGE_SIZE) < m1 * EB_PAGE_SIZE, n < m1 * EB_PAGE_SIZE, n0 + # buf1 < EB_PAGE_SIZE, n1 < EB_PAGE_SIZE, n0 < EB_PAGE_SIZE, (n0 = 0 ⊃ # buf0;m0 * EB_PAGE_SIZE) ≤ n ⊦ m1 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; (ws)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ EB_PAGE_SIZE ≠ 0; (s)