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Proof step 18 from proof gcdnex-prime (previous step)

2 ≤ n0, m1 ≤ n0, prime(m1), 2 ≤ m1, n0 * m1 | m0, n0 * m1 | m ⊦ n0 = 0, n0 * m1 = 1, m1 = 0, m0 = 1, m = 1, prime(m1) ∧ m1 ≠ 0 ∧ m1 ≠ 1 ∧ m1 | m ∧ m1 | m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
n0 * m1 | m, n0 * m1 | m0, 2 ≤ m1, prime(m1), m1 ≤ n0, 2 ≤ n0 ⊦ m = 1, m0 = 1, m1 = 0, n0 * m1 = 1, n0 = 0, m1 = 1, m1 | m0 ∧ m1 | m

Rule argument

Additional information

Used simplifier rules: one: ⊦ ¬ prime(1);%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%one: ⊦ m0 * 1 = m0; from specification nat-mult