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Proof step 9 from proof Binom-theorem-countFun (previous step)

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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
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)) ⊦ (i + i0) ^ i1 * (i + i0) = Σ countFun(0, i1 + 1, λ j0. binom_coef(i1 + 1, j0) * i ^ j0 * i0 ^ (i1 + 1 - j0)), ~1 = i1

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs