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Proof step 26 from proof gcd-mod-one (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mod-one-01: ⊦ n ≠ 0 ∧ n ≠ 1 → 1 % n = 1; from specification nat-div%mod-mult: n ≠ 0 ⊦ (m + n0 * n) % n = m % n; from specification nat-div