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Proof step 5 from proof modls-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m0 * n + n1 < (m - 1) * n + n2, m0 + 1 < m, n1 < n, m * n + n2 ≤ (n * m0 + n) + n1, n2 < n ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%ls: n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%pred-03: ⊦ n -1 = n - 1; from specification nat