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Proof step 5 from proof le-div-to-mult (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 6:
n * m0 ≤ m1 * m0 + n0, n0 < m0 ⊦ n ≤ m1, m0 = 0

Rule argument

pos: right 1

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%modls: ⊦ n0 < m0 → (m0 * m + n0 < m0 * n ↔ m < n); from specification nat-mult%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2