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Proof step 3 from proof ls-01 (previous step)

∀ m0. m0 * m1 < m1 → m1 = 0 ∨ m0 = 0, m0 * m1 +1 < m1 +1 ⊦ m1 +1 = 0, m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m1 * m0 + m0 ≤ m1, ∀ m0. m0 * m1 < m1 → m1 = 0 ∨ m0 = 0 ⊦ m0 = 0

Rule argument

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2