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Proof step 11 from proof mod-eq-01 (previous step)

m0 + n0 = m1 * n + n1, n1 < n, m0 < n, n0 < n ⊦ n1 = m0 ↔ n0 = 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m = m + n ↔ n = 0; (s)% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; (ws)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m * n + n0 < n ↔ m = 0 ∧ n0 < n; (ws)% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 0; (ws)