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Proof step 1 from proof coef-22 (conclusion)

⊦ n ≤ m1 * n + m2 * n ↔ ¬ ¬(1 ≤ m1 + m2 ∨ n = 0)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%coef-16: ⊦ m * n + m0 * n < n ↔ ¬ ¬(m + m0 = 0 ∧ n ≠ 0);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat