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Proof step 5 from proof mod-zero (previous step)

n < m, 2 < m, prime(m), 2 ≤ n, ∃ n0. m = n0 * n ⊦ m = 2, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
2 < n0 * n, prime(n0 * n), 2 ≤ n ⊦ n0 * n = 2, n = 0, n0 = 0, n0 = 1

Rule argument

Additional information

Used simplifier rules: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; (ws)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)% ⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m); (s)