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Proof step 4 from proof coef-02 (previous step)

n * (m + m0) ≤ n * (m1 + m2) ↔ m + m0 ≤ m1 + m2 ⊦ m * n + m0 * n ≤ m1 * n + m2 * n ↔ m + m0 ≤ m1 + m2 ∨ n = 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

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