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Proof step 16 from proof modprime (previous step)

∀ m1. 2 ≤ m1 ∧ m1 < m0 * m → (m0 * m) % m1 ≠ 0 ⊦ m0 * m = 1, m = 1, m = 0, m0 = 0, m0 = 1

User applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: m0 abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m); (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%n ≠ 0 ⊦ (m * n) % n = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m ≠ 1 → (m < 2 ↔ m = 0); (s) Used terms: m0