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Proof step 8 from proof modprime (previous step)

n = m0 * m + n0 ∧ n0 < m, n0 = 0, 2 ≤ m, m < n, 2 < n, ∀ m. m | n → m = 1 ∨ m = n ⊦ n = 1, n = 0, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
2 < m0 * m, 2 ≤ m, ∀ m1. m1 | m0 * m → m1 = 1 ∨ m1 = m0 * m ⊦ m0 = 1, m0 = 0, m0 * m = 1, m = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%ls-04: ⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m); from specification nat-mult%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2