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Proof step 1 from proof max_exp-not-div (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
prime(i), 0 < j0, 0 < j ⊦ max_exp(i, j * j0) = max_exp(i, j0), i | j

Rule argument

Additional information

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