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Proof step 3 from proof leone (previous step)

∀ m. (m * n0 ≤ n0 ↔ m ≤ 1) ∨ n0 = 0 ⊦ m * n0 +1 ≤ n0 +1 ↔ m ≤ 1, n0 +1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ m. m * n0 ≤ n0 ↔ m ≤ 1 ⊦ n0 * m + m ≤ n0 + 1 ↔ m ≤ 1, n0 = 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;%zero: ⊦ 0 * n = 0;%add-base: ⊦ n + 0 = n; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2