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Proof step 3 from proof max_prime_divisor-divides-01 (previous step)

0 < i, 2 ≤ i, 0 ≤ i ⊦ i % max_prime_divisor(i) = 0, prime(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ j | i ∧ i ≠ 0 → i % j = 0

Rule argument

spec: <toplevel> name: div-mod-zero seq: ⊦ j | i ∧ i ≠ 0 → i % j = 0 subst: [j, i] → [max_prime_divisor(i), i] flags: <all paths>

Additional information

Used simplifier rules: max_prime_divisor-divides: ⊦ ¬ prime(i) ∧ 2 ≤ i → max_prime_divisor(i) | i; Used terms: max_prime_divisor(i), i