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Proof step 9 from proof modprime (previous step)

2 < m0 * m, 2 ≤ m, ∀ m1. m1 | m0 * m → m1 = 1 ∨ m1 = m0 * m ⊦ m0 * m = 1, m = 0, m0 = 0, m0 = 1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: m0

Additional information

Used simplifier rules: divi-mult: ⊦ m | m * n0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m * n = m ↔ m = 0 ∨ n = 1; (s)% ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1; (ws) Used terms: m0