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Proof step 11 from proof max_exp-not-div (previous step)

prime(i), 0 < j, i | j0, 0 < j0, ∀ i0, j1, j. i→n(max_exp(i0, j1)) < i→n(max_exp(i, j0 / i) + 1) → prime(i0) ∧ 0 < j1 ∧ 0 < j → max_exp(i0, j * j1) = max_exp(i0, j1) ∨ i0 | j ⊦ max_exp(i, (j * j0) / i) = max_exp(i, j0 / i), i | j

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [i0, j1, j] → [i, j0 / i, j]

Additional information

Used simplifier rules: max_exp-pos-04: ⊦ ¬ max_exp(i, j) < 0;%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%abs-le-succ: ⊦ abs(i + 1) ≤ abs(i) ↔ i < 0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i, j0 / i, j