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

0 < i, prime(i) ⊦ max_exp(i, i * fac(i - 1)) = 1

User applied rule insert rewrite lemma

And obtained the following new goals

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

Rule argument

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

Additional information

Used simplifier rules: fac-pos-01: ⊦ ¬ fac(i) ≤ 0;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs Used terms: fac(i - 1), i, i