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Proof step 3 from proof max_exp-fac-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 < i, 0 ≤ j0, j0 < i, prime(i), ∀ j, i. i→n(j) < i→n(j0) → prime(i) ∧ j < i ∧ 0 ≤ j ∧ 0 < i → max_exp(i, fac(i + j)) = 1 ⊦ max_exp(i, fac(i + j0)) = 1

Proof step 3 from proof max_exp-fac-01 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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