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Proof step 19 from proof max_exp-mult-01 (previous step)

¬ i | j ∧ prime(i) ∧ 0 < j, 0 < j0, 0 < j, prime(i), ∀ i0, j1, j0. i→n(max_exp(i0, j1)) < i→n(0) → prime(i0) ∧ 0 < j1 ∧ 0 < j0 → max_exp(i0, j1 * j0) = max_exp(i0, j1) + max_exp(i0, j0) ⊦ i | j, max_exp(i, j * j0) = 0 + max_exp(i, j0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ax-04: ⊦ i + 0 = i; from specification int-basic2