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Proof step 6 from proof max_exp-fac-01 (previous step)

j0 < i, prime(i), 1 < i, 0 < i, 0 < j0, ∀ 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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ 0 < i → fac(i) = i * fac(i - 1)
Proof step 8:
0 < i + j0, j0 < i, prime(i), 1 < i, 0 < i, 0 < j0, ∀ 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, (i + j0) * fac(i + j0 - 1)) = 1

Rule argument

spec: inst: name: fac-rec seq: ⊦ 0 < i → fac(i) = i * fac(i - 1) subst: [i] → [i + j0] flags: rewrite, current

Additional information

Used simplifier rules: zero-02: ⊦ 0 < i ∧ 0 < j → 0 < i + j; from specification int-basic2%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