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Proof step 16 from proof max_exp-fac-rec (previous step)

0 < j ∧ 0 < fac(j - 1) ∧ prime(i), 0 < i, 0 ≤ j0, j0 < i, prime(i), 0 < j, 0 < fac(i * 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 * 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 17:
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 * j) + max_exp(i, fac(i * j - 1)) = j + max_exp(i, j) + max_exp(i, fac(j - 1))

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs