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Proof step 14 from proof gcd-conq-one (previous step)

m * m0 = m1 * n + 1, gcd(m, n) = 1 ⊦ n * m1 = dist(m * m0, 1), n = 1, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ n ≤ m → dist(m, n) = m - n
Proof step 16:
m * m0 = m1 * n + 1, gcd(m, n) = 1, 1 ≤ m * m0 ⊦ n * m1 = m * m0 - 1, n = 0, n = 1

Rule argument

spec: <toplevel> name: dist-snd-01 seq: ⊦ n ≤ m → dist(m, n) = m - n subst: [n, m] → [1, m * m0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: 1, m * m0