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Proof step 4 from proof fac-pos-01 (previous step)

fac(i) ≤ 0, ∀ i0. i→n(i0) < i→n(i) → fac(i0) ≤ 0 → false ⊦

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

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