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Proof step 10 from proof Binom-theorem-countFun (previous step)

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

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [j0, i, i0] → [i1, i, i0]

Additional information

Used simplifier rules: rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1; from specification intnat%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%irreflexivity: ⊦ ¬ i < i; from specification oelem[intlista]%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: i1, i, i0