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Proof step 14 from proof countFun-pot (previous step)

0 < i1 + 1 - i2, i2 ≤ i1, ∀ i3, i4, i0, i. i→n(i3 - i4) < i→n(i1 - i2) → true → Σ countFun(i4, i3, λ j0. i0 * binom_coef(i3, j0) * i ^ j0 * i0 ^ (i3 - j0)) = Σ countFun(i4, i3, λ j0. binom_coef(i3, j0) * i ^ j0 * i0 ^ (i3 + 1 - j0)) ⊦ i0 * binom_coef(i1, i2) * i ^ i2 * i0 ^ (i1 - i2) + Σ countFun(i2 + 1, i1, λ j0. i0 * binom_coef(i1, j0) * i ^ j0 * i0 ^ (i1 - j0)) = binom_coef(i1, i2) * i ^ i2 * i0 ^ (i1 + 1 - i2)-1 * i0 + Σ countFun(i2 + 1, i1, λ j0. binom_coef(i1, j0) * i ^ j0 * i0 ^ (i1 + 1 - j0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%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%sub-add-06: ⊦ i + j - i0 + j = i - i0; from specification int-abs%sub-sub: ⊦ i - j - i0 = i - j + i0; from specification int-abs%sub-pred: ⊦ i -1 = i - 1; from specification int-abs