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Proof step 18 from proof content-trunc (previous step)

n1 < n2 + m, m0 * (n2 + m) < n0 * (n2 + m) + n1, m0 ∈ content, (n2 + m) + (n2 + m) * n0 ≤ n2 + m0 * (n2 + m) ⊦ content[m0][n2] = zero, m = 0, n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
m < n1, m0 * (n2 + m) < n0 * (n2 + m) + n1, n1 < n2 + m, m0 ∈ content, m + (n2 + m) * n0 ≤ m0 * (n2 + m) ⊦ content[m0][n2] = zero, m = 0, n2 = 1, n2 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)% ⊦ m +1 = m + 1; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)