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Proof step 14 from proof fac-add (previous step)

0 ≤ i, 0 < j, 0 < i + 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 - 1) = fac(i) * Π countFun(1, j - 1, λ j0. i + j0)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [j0, i] → [j - 1, i]

Additional information

Used simplifier rules: sub-add-01: ⊦ i + (j - i0) = i + j - i0; 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%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: j - 1, i