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Proof step 5 from proof countFun-shift (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i2 < i1, ∀ i3, i4, i, i0. i→n(i3 - 1 - i4) < i→n(i1 - 1 - i2) → true → Σ countFun(i4, i3 - 1, λ j0. i * binom_coef(i3, j0) * i ^ j0 * i0 ^ (i3 - j0)) = Σ countFun(i4 + 1, i3, λ j0. i * binom_coef(i3, j0 - 1) * i ^ (j0 - 1) * i0 ^ (i3 - (j0 - 1))) ⊦ Σ countFun(i2, i1 - 1, λ j0. i * binom_coef(i1, j0) * i ^ j0 * i0 ^ (i1 - j0)) = Σ countFun(i2 + 1, i1, λ j. i * binom_coef(i1, j - 1) * i ^ (j - 1) * i0 ^ (i1 + 1 - j))

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs