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Proof step 5 from proof modp (previous step)

n0 < m0, m1 + 1 < n, m0 * n ≤ (m0 * m1 + m0) + n0, ¬ m1 < n -1 ∨ ¬ m0 * n -1 ≤ m0 * m1 + n0 ∨ ¬ n0 < m0, ∀ n1, m, n2. ¬ m1 < n1 ∨ ¬ m * n1 ≤ m * m1 + n2 ∨ ¬ n2 < m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m0 * m1 + n0 < m0 * (n - 1), m0 * n ≤ (m0 * m1 + m0) + n0, m1 + 1 < n, n0 < m0, ∀ n1, m, n2. ¬ m1 < n1 ∨ ¬ m * n1 ≤ m * m1 + n2 ∨ ¬ n2 < m ⊦ n - 1 = 0

Rule argument

Additional information

Used simplifier rules: zero-01: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%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