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Proof step 2 from proof max_exp-not-div (previous step)

prime(i), 0 < j0, 0 < j ⊦ max_exp(i, j * j0) = max_exp(i, j0), i | j

User applied rule induction

And obtained the following new goals

Proof step 3:
prime(i) ∧ 0 < j0 ∧ 0 < j, ∀ i0, j1, j. i→n(max_exp(i0, j1)) < i→n(max_exp(i, j0)) → prime(i0) ∧ 0 < j1 ∧ 0 < j → max_exp(i0, j * j1) = max_exp(i0, j1) ∨ i0 | j ⊦ max_exp(i, j * j0) = max_exp(i, j0) ∨ i | j

Rule argument

indvar: i→n(max_exp(i, j0)) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(max_exp(i, j0))