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Proof step 5 from proof max_exp-fac-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < i, 0 ≤ j0, 0 < j, j0 < i, prime(i), ∀ 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, fac(i * j)) = j + max_exp(i, fac(j))

Proof step 5 from proof max_exp-fac-rec (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information