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Proof step 4 from proof max_exp-fac-smaller (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < i → fac(i) = i * fac(i - 1)
Proof step 6:
0 < j, 1 < i, 0 < i, 0 < j, j < i, prime(i), ∀ j0, i. i→n(j0) < i→n(j) → prime(i) ∧ j0 < i ∧ 0 < j0 ∧ 0 < i ∧ 1 < i → max_exp(i, fac(j0)) = 0 ⊦ max_exp(i, j * fac(j - 1)) = 0

Rule argument

spec: inst: name: fac-rec seq: ⊦ 0 < i → fac(i) = i * fac(i - 1) subst: [i] → [j] flags: rewrite, current

Additional information

Used terms: j