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Proof step 4 from proof max_exp-mult-01 (previous step)

0 < j0, 0 < j, prime(i), ∀ i0, j1, j0. i→n(max_exp(i0, j1)) < i→n(max_exp(i, j)) → prime(i0) ∧ 0 < j1 ∧ 0 < j0 → max_exp(i0, j1 * j0) = max_exp(i0, j1) + max_exp(i0, j0) ⊦ max_exp(i, j * j0) = max_exp(i, j) + max_exp(i, j0)

User applied rule cut formula

And obtained the following new goals

Proof step 5:
i | j, 0 < j0, 0 < j, prime(i), ∀ i0, j1, j0. i→n(max_exp(i0, j1)) < i→n(max_exp(i, j)) → prime(i0) ∧ 0 < j1 ∧ 0 < j0 → max_exp(i0, j1 * j0) = max_exp(i0, j1) + max_exp(i0, j0) ⊦ max_exp(i, j * j0) = max_exp(i, j) + max_exp(i, j0)
Proof step 17:
0 < j0, 0 < j, prime(i), ∀ i0, j1, j0. i→n(max_exp(i0, j1)) < i→n(max_exp(i, j)) → prime(i0) ∧ 0 < j1 ∧ 0 < j0 → max_exp(i0, j1 * j0) = max_exp(i0, j1) + max_exp(i0, j0) ⊦ i | j, max_exp(i, j * j0) = max_exp(i, j) + max_exp(i, j0)

Rule argument

fma: i | j

Additional information

Used formulas: i | j