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Proof step 6 from proof modls-03 (previous step)

m0 + 1 < m, n2 < n, n1 < n, m * n + n2 ≤ (n * m0 + n) + n1, ¬ n2 < n ∨ ¬ m0 < m - 1 ∨ ¬ n1 < n ∨ ¬ (m - 1) * n + n2 ≤ m0 * n + n1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m0 * n + n1 < (m - 1) * n + n2, m0 + 1 < m, n2 < n, n1 < n, m * n + n2 ≤ (n * m0 + n) + n1 ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)