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Proof step 5 from proof Binom-theorem-countFun (previous step)

0 < j, ∀ j0, i, i0. i→n(j0) < i→n(j) → true → 0 ≤ j0 → (i + i0) ^ j0 = Σ countFun(0, j0, λ j1. binom_coef(j0, j1) * i ^ j1 * i0 ^ (j0 - j1)) ⊦ j = 0, (i + i0) ^ j = Σ countFun(0, j, λ j0. binom_coef(j, j0) * i ^ j0 * i0 ^ (j - j0))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 7:
0 < j, 0 < j, ∀ j0, i, i0. i→n(j0) < i→n(j) → true → 0 ≤ j0 → (i + i0) ^ j0 = Σ countFun(0, j0, λ j1. binom_coef(j0, j1) * i ^ j1 * i0 ^ (j0 - j1)) ⊦ j = 0, (i + i0) ^ j -1 * (i + i0) = Σ countFun(0, j, λ j0. binom_coef(j, j0) * i ^ j0 * i0 ^ (j - j0))

Rule argument

spec: int-pot inst: name: Pot-pos seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [j, i + i0] flags: rewrite

Additional information

Used terms: j, i + i0