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Proof step 23 from proof gcd-mod-one (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 3 flags: rotate, discard

Additional information

Used simplifier rules: rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%z: ⊦ 0 + m = m; from specification nat-basic2%eq: ⊦ m0 ≠ 0 → (m * m0 = n * m0 ↔ m = n); from specification nat-mult