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Proof step 16 from proof not-prime (previous step)

n0 < m, 2 < m * m, 2 ≤ m * n0, ∀ n1. n1 < m * n0 → ¬ prime(n1) ∧ 2 ≤ n1 → (∃ m. m | n1 ∧ prime(m) ∧ m * m ≤ n1) ⊦ n0 = 0, m = 1, n0 = 1, m * n0 = 1, m = 0, ∃ m0. m0 | m * n0 ∧ prime(m0) ∧ m0 * m0 ≤ m * n0

User applied rule exists right

And obtained the following new goals

Proof step 17:
n0 < m, 2 < m * m, 2 ≤ m * n0, ∀ n1. n1 < m * n0 → ¬ prime(n1) ∧ 2 ≤ n1 → (∃ m. m | n1 ∧ prime(m) ∧ m * m ≤ n1) ⊦ n0 = 0, m = 1, n0 = 1, m * n0 = 1, m = 0, n0 | m * n0 ∧ prime(n0) ∧ n0 * n0 ≤ m * n0, ∃ m0. m0 | m * n0 ∧ prime(m0) ∧ m0 * m0 ≤ m * n0

Rule argument

pos: right 6 terms: n0 abort: F discard: F computed: F

Additional information

Used terms: n0