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Proof step 19 from proof gcd-mod-one (previous step)

m0 * m = n * n0 - 1, m0 ≤ n * n0, gcd(m, n) = 1, m * m0 + 1 = n * n0, (n * n0) * m - (n * n0 - 1) = m1 * n + n1, m0 ≤ n * n0, n1 < n ⊦ n * (n0 * m - n0) + 1 = (n * n0) * m - (n * n0 - 1), n1 = 1, n = 1, m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ m0 ≤ m → (m - m0) * n = m * n - m0 * n
Proof step 21:
n0 ≤ n0 * m, m0 * m = n * n0 - 1, m0 ≤ n * n0, gcd(m, n) = 1, m * m0 + 1 = n * n0, (n * n0) * m - (n * n0 - 1) = m1 * n + n1, m0 ≤ n * n0, n1 < n ⊦ ((n0 * m) * n - n0 * n) + 1 = (n * n0) * m - (n * n0 - 1), n1 = 1, n = 1, m = 0

Rule argument

spec: nat-mult inst: name: dist-sub seq: ⊦ m0 ≤ m → (m - m0) * n = m * n - m0 * n subst: [n, m0, m] → [n, n0, n0 * m] flags: rewrite

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%ls-01: ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 0; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: n, n0, n0 * m