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Proof step 9 from proof max_prime_divisor-ge2 (previous step)

max_prime_divisor(i) * i0 = i, (max_prime_divisor(i) * i0) / max_prime_divisor(i) < 2, 0 < i, (max_prime_divisor(i) * i0) / max_prime_divisor(i) < i, 2 ≤ i, 0 ≤ i ⊦ prime(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
j ≠ 0 ⊦ (j * i) / j = i
Proof step 11:
max_prime_divisor(i) * i0 = i, i0 < i, 0 < i, i0 < 2, 0 ≤ i, 2 ≤ i ⊦ max_prime_divisor(i) = 0, prime(i)

Rule argument

spec: int-div name: div-mult seq: j ≠ 0 ⊦ (j * i) / j = i subst: [i, j] → [i0, max_prime_divisor(i)] flags: <all paths>

Additional information

Used simplifier rules: max_prime_divisor-not-zero: ⊦ ¬ prime(i) ∧ 2 ≤ i → max_prime_divisor(i) ≠ 0; Used terms: i0, max_prime_divisor(i)