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Proof step 15 from proof three (previous step)

m = m0 +1, m1 * m + m ≤ 3 ⊦ m = 0, m = 1, 1 = m1, m = 2, m1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
m0 * m1 + m1 + m0 ≤ 2 ⊦ m1 = 0, m0 = 0, 1 = m1, 1 = m0

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;%le-num-05: n < n0 ⊦ m + (* n) ≤ (* n0) ↔ m ≤ n0 - n; from specification nat%add-num-04: n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; from specification nat%one: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; 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