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Proof step 9 from proof fac-add (previous step)

0 ≤ i, 0 < j, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 0 ≤ j0 → fac(i + j0) = fac(i) * Π countFun(1, j0, λ j0. i + j0) ⊦ j = 0, fac(i + j) = fac(i) * Π countFun(1, j, λ j0. i + j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: fac-zero: ⊦ i ≤ 0 → fac(i) = 1;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%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 + j