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Proof step 24 from proof Binom-theorem-countFun (previous step)

¬(0 < 1 ∧ 1 ≤ i1), 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)) + i ^ (i1 + 1)

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;%eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%one: ⊦ i ^ 1 = i; from specification int-pot%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%one: ⊦ 1 * i = i; from specification int-mult%sum-one: ⊦ Σ i ' = i; from specification intlist%commutativity: ⊦ i * j = j * i; from specification int-mult%Pot-zero: ⊦ j ^ 0 = 1; from specification int-pot%countFun-one: ⊦ countFun(i, i, intfun) = intfun(i) '; from specification intlist%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs