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Proof step 3 from proof numeral-10 (previous step)

m1 * m + n0 < m2 * m, m2 < m1 ⊦ m = 0, m2 = 0

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
m2 < m0 +1, m0 +1 * m + n0 < m2 * m, ∀ m, n0, m2. m0 * m + n0 < m2 * m ∧ m2 < m0 → m = 0 ∨ m2 = 0 ⊦ m = 0, m2 = 0

Rule argument

term: m1

Additional information

Used simplifier rules: mult-base: ⊦ m * 0 = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%nat freely generated by 0, +1;;% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ¬ m < 0; (s)