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Proof step 5 from proof aligndown-add-not-smaller (previous step)

(n0 + n1) % m = 0, n0 + n1 < (n0 % m = 0 ⊃ n0;n0 / m * m) ⊦ m = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
(n0 + n1) % m = 0, n0 + n1 < n0 / m * m ⊦ n0 % m = 0, m = 0

Rule argument

pos: left 2

Additional information

Used simplifier rules: ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) % n = m % n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ¬ m + n < m; (s)