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Proof step 4 from proof fac-le (previous step)

0 < j, 0 < i, ∀ i0, j. i→n(i0) < i→n(i) → 0 < i0 ∧ 0 < j → (fac(i0) ≤ fac(j) ↔ i0 ≤ j) ⊦ fac(i) ≤ fac(j) ↔ i ≤ j

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 < i, 0 < j, 0 < i, ∀ i0, j. i→n(i0) < i→n(i) → 0 < i0 ∧ 0 < j → (fac(i0) ≤ fac(j) ↔ i0 ≤ j) ⊦ i * fac(i - 1) ≤ fac(j) ↔ i ≤ j

Rule argument

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

Additional information

Used terms: i