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Proof step 21 from proof Binom-theorem-countFun (previous step)

0 ≤ i1 + 1, 0 ≤ i1, ∀ j0, i, i0. i→n(j0) < i→n(i1 + 1) → true → 0 ≤ j0 → (i + i0) ^ j0 = Σ countFun(0, j0, λ j1. binom_coef(j0, j1) * i ^ j1 * i0 ^ (j0 - j1)) ⊦ ~1 = i1, Σ countFun(0, i1, λ j1. binom_coef(i1, j1) * i ^ j1 * i0 ^ (i1 - j1)) * (i + i0) = i0 ^ (1 + i1) + Σ countFun(1, i1, λ j0. binom_coef(i1 + 1, j0) * i ^ j0 * i0 ^ (i1 + 1 - j0)) + 1 * i ^ (i1 + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult