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Proof step 4 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 = 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 5:
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 = Σ countFun(0, j, λ j0. binom_coef(j, j0) * i ^ j0 * i0 ^ (j - j0))

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs