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Proof step 11 from proof not-prime (previous step)

m0 | m, prime(m0), m0 * m0 ≤ m, m * m ≤ n, 2 ≤ n, m | n, m0 | n, ∀ n0. n0 < n → ¬ prime(n0) ∧ 2 ≤ n0 → (∃ m. m | n0 ∧ prime(m) ∧ m * m ≤ n0) ⊦ n = 0, m = 1, m = n, n = 1, prime(m), ∃ m. m | n ∧ prime(m) ∧ m * m ≤ n

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 6 terms: m0

Additional information

Used simplifier rules: divi-ls: ⊦ m < n0 → (n0 | m ↔ m = 0);%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%sqzero: ⊦ m * m = 0 ↔ m = 0; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%square-less: ⊦ m * m < n * n ↔ m < n; from specification nat-mult Used terms: m0