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

⊦ ∃ i, j, j0. 0 < j ∧ 0 < j0 ∧ prime(i) ∧ ¬ j0 | j ∧ max_exp(i, j / j0) ≠ max_exp(i, j) - max_exp(i, j0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ∃ i, j, j0. 0 < j ∧ 0 < j0 ∧ prime(i) ∧ ¬ j0 | j ∧ max_exp(i, j) ≠ max_exp(i, j / j0) + max_exp(i, j0)

Rule argument

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs