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Proof step 20 from proof max_exp-fac-rec (previous step)

0 < i, 0 ≤ j0, j0 < i, prime(i), 0 < j, 0 < fac(i * j - 1), 0 < fac(j - 1), ∀ j1, i, j0. i→n(j1) < i→n(j) → true → prime(i) ∧ 0 < j1 ∧ 0 ≤ j0 ∧ j0 < i → max_exp(i, fac(i * j1 + j0)) = j1 + max_exp(i, fac(j1)) ⊦ 1 + max_exp(i, fac(i * j - 1)) = j + max_exp(i, fac(j - 1))

User applied rule cut formula

And obtained the following new goals

Proof step 21:
i * j - 1 = i * (j - 1) + (i - 1), 0 < i, 0 ≤ j0, j0 < i, prime(i), 0 < j, 0 < fac(i * j - 1), 0 < fac(j - 1), ∀ j1, i, j0. i→n(j1) < i→n(j) → true → prime(i) ∧ 0 < j1 ∧ 0 ≤ j0 ∧ j0 < i → max_exp(i, fac(i * j1 + j0)) = j1 + max_exp(i, fac(j1)) ⊦ 1 + max_exp(i, fac(i * j - 1)) = j + max_exp(i, fac(j - 1))
Proof step 27:
0 < i, 0 ≤ j0, j0 < i, prime(i), 0 < j, 0 < fac(i * j - 1), 0 < fac(j - 1), ∀ j1, i, j0. i→n(j1) < i→n(j) → true → prime(i) ∧ 0 < j1 ∧ 0 ≤ j0 ∧ j0 < i → max_exp(i, fac(i * j1 + j0)) = j1 + max_exp(i, fac(j1)) ⊦ i * j - 1 = i * (j - 1) + (i - 1), 1 + max_exp(i, fac(i * j - 1)) = j + max_exp(i, fac(j - 1))

Rule argument

fma: i * j - 1 = i * (j - 1) + (i - 1)

Additional information

Used formulas: i * j - 1 = i * (j - 1) + (i - 1)