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Proof step 14 from proof prime-smallest (previous step)

prime(m), m ≤ m0, ∀ m1. 2 ≤ m1 ∧ m1 * m1 ≤ m0 * m → (m0 * m) % m1 ≠ 0 ⊦ m = 0, m0 * m = 1, m0 = 0, m0 = 1

User applied rule all left

And obtained the following new goals

Proof step 15:
prime(m), m ≤ m0, 2 ≤ m0 ∧ m0 * m0 ≤ m0 * m → (m0 * m) % m0 ≠ 0, ∀ m1. 2 ≤ m1 ∧ m1 * m1 ≤ m0 * m → (m0 * m) % m1 ≠ 0 ⊦ m = 0, m0 * m = 1, m0 = 0, m0 = 1

Rule argument

pos: left 3 terms: m0 abort: F discard: F computed: F

Additional information

Used terms: m0