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Proof step 16 from proof three (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
m1 * m0 + m1 + m0 ≤ 2 ⊦ 1 = m1, 1 = m0, m0 = 0, m1 = 0

Rule argument

Additional information

Used simplifier rules: rsucc-02: ⊦ m * (n + 1) = m * n + m;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%n < n0 ⊦ m + (* n) ≤ (* n0) ↔ m ≤ n0 - n; (s)% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws)% ⊦ m + 1 ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; (s)% ⊦ m +1 = m + 1; (s)