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Proof step 23 from proof max_exp-fac-rec (previous step)

prime(i) ∧ 0 < j - 1 ∧ 0 ≤ i - 1 ∧ i - 1 < i → max_exp(i, fac(i * (j - 1) + (i - 1))) = (j - 1) + max_exp(i, fac(j - 1)), 0 < i, 0 ≤ j0, j0 < i, prime(i), 0 < j, 0 < fac(i * (j - 1) + (i - 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) + (i - 1))) = j + max_exp(i, fac(j - 1))

User applied rule case distinction

And obtained the following new goals

Proof step 24:
0 < i, 0 ≤ j0, j0 < i, prime(i), 0 < j, 0 < fac(i * (j - 1) + (i - 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)) ⊦ prime(i) ∧ 0 < j - 1 ∧ 0 ≤ i - 1 ∧ i - 1 < i, 1 + max_exp(i, fac(i * (j - 1) + (i - 1))) = j + max_exp(i, fac(j - 1))

Rule argument

pos: left 1

Additional information

Used simplifier rules: sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs