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Proof step 19 from proof max_exp-fac-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
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))

Rule argument

Additional information

Used simplifier rules: max_exp-same: ⊦ prime(i) → max_exp(i, i) = 1;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2