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Proof step 15 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)) ⊦ Σ countFun(0, i1, λ j1. binom_coef(i1, j1) * i ^ j1 * i0 ^ (i1 - j1)) * (i + i0) = Σ((binom_coef(i1 + 1, 0) * i ^ 0 * i0 ^ (i1 + 1 - 0))' + countFun(0 + 1, i1 + 1, λ j0. binom_coef(i1 + 1, j0) * i ^ j0 * i0 ^ (i1 + 1 - j0))), ~1 = i1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: coef-zero: ⊦ 0 ≤ i → binom_coef(i, 0) = 1;%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%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%sum-rec: ⊦ Σ(i ' + ints) = i + Σ ints; from specification intlist%one: ⊦ 1 * i = i; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%Pot-zero: ⊦ j ^ 0 = 1; from specification int-pot