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Proof step 4 from proof align-down-add-eq-add-ls (previous step)

((n0 + n1) % m = 0 ⊃ n0 + n1;(n0 + n1) / m * m) = n2 + n1, n0 < n2 ⊦ m = 0

User applied rule case distinction

And obtained the following new goals

Proof step 5:
(n0 + n1) / m * m = n2 + n1, n0 < n2 ⊦ (n0 + n1) % m = 0, m = 0

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s)