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Proof step 8 from proof gcd-mod-one (previous step)

m * m0 + 1 = n * n0, gcd(m, n) = 1 ⊦ (m * (n * n0 - m0)) % n = 1, m = 0, n = 1, n = 0

User applied rule cut formula

And obtained the following new goals

Proof step 9:
m * m0 + 1 = n * n0, gcd(m, n) = 1 ⊦ (m * (n * n0 - m0)) % n = 1, m = 0, n = 0, n = 1, n * n0 < m0

Rule argument

fma: n * n0 < m0

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m * n + n0 < n ↔ m = 0 ∧ n0 < n; (ws)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: n * n0 < m0