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Proof step 13 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)) ⊦ Σ countFun(0, i1, λ j1. binom_coef(i1, j1) * i ^ j1 * i0 ^ (i1 - j1)) * (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 insert rewrite lemma

And obtained the following new goals

Rule argument

spec: intlist inst: name: countFun-rec seq: ⊦ i ≤ j → countFun(i, j, intfun) = intfun(i) ' + countFun(i + 1, j, intfun) subst: [intfun, j, i] → [λ j0. binom_coef(i1 + 1, j0) * i ^ j0 * i0 ^ (i1 + 1 - j0), i1 + 1, 0] flags: rewrite

Additional information

Used simplifier rules: numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: λ j0. binom_coef(i1 + 1, j0) * i ^ j0 * i0 ^ (i1 + 1 - j0), i1 + 1, 0