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Proof step 12 from proof max_exp-fac-01 (previous step)

j0 < i, prime(i), 1 < i, 0 < i, 0 < j0, 0 < fac(i + j0 - 1), ∀ j, i. i→n(j) < i→n(j0) → prime(i) ∧ j < i ∧ 0 ≤ j ∧ 0 < i → max_exp(i, fac(i + j)) = 1 ⊦ j0 = 0, max_exp(i, fac(i + j0 - 1)) = 1

User applied rule apply induction

And closed the goal!

Rule argument

subst: [j, i] → [j0 - 1, i]

Additional information

Used simplifier rules: sub-add-01: ⊦ i + (j - i0) = i + j - i0; 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%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs Used terms: j0 - 1, i