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Proof step 6 from proof modprime (previous step)

∀ m. m | n → m = 1 ∨ m = n ⊦ ∀ m. 2 ≤ m ∧ m < n → n % m ≠ 0, n = 1, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n % m = 0, 2 ≤ m, m < n, 2 < n, ∀ m. m | n → m = 1 ∨ m = n ⊦ n = 1, n = 0, m = 0

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1