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Proof step 8 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 - 1) * (i + i0) = Σ countFun(0, j, λ j0. binom_coef(j, j0) * i ^ j0 * i0 ^ (j - j0))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
j = i1 + 1, 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) ^ i1 * (i + i0) = Σ countFun(0, j, λ j0. binom_coef(j, j0) * i ^ j0 * i0 ^ (j - j0))

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, 1, i1] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 1, i1