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Proof step 15 from proof countFun-pot-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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%abs-03: ⊦ 0 ≤ abs(i); from specification int-abs%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%abssub: ⊦ j ≤ i → abs(i - j) = i - j; from specification int-abs%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat