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Proof step 12 from proof three (previous step)

∀ m. m * m1 ≤ 3 → m1 = 2 ∨ m = 2 ∨ m1 = 1 ∨ m1 = 0 ∨ m = 0 ∨ m = 1, m * m1 +1 ≤ 3 ⊦ m1 +1 = 2, m = 2, m1 +1 = 1, m1 +1 = 0, m = 0, m = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
m1 * m + m ≤ 3, ∀ m0. m0 * m1 ≤ 3 → m1 = 2 ∨ m0 = 2 ∨ m0 = 0 ∨ m0 = 1 ⊦ m = 0, m = 1, 1 = m1, m = 2, m1 = 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;%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%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%add-num-04: n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2