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

j1 ≤ i1, 0 < j1, 0 < i1, ∀ i2, j, i, i0. i→n(i2 - j) < i→n(i1 - j1) → true → 0 < j ∧ j ≤ i2 → countFun(j, i2, λ j0. binom_coef(i2 + 1, j0) * i ^ j0 * i0 ^ (i2 + 1 - j0)) = countFun(j, i2, λ j0. binom_coef(i2, j0 - 1) * i ^ j0 * i0 ^ (i2 + 1 - j0) + binom_coef(i2, j0) * i ^ j0 * i0 ^ (i2 + 1 - j0)) ⊦ i→n(i1 - j1 + 1) < i→n(i1 - j1), ((binom_coef(i1, j1 - 1) + binom_coef(i1, j1)) * i ^ j1 * i0 ^ (i1 + 1 - j1))' + countFun(j1 + 1, i1, λ j0. binom_coef(i1 + 1, j0) * i ^ j0 * i0 ^ (i1 + 1 - j0)) = (binom_coef(i1, j1 - 1) * i ^ j1 * i0 ^ (i1 + 1 - j1) + binom_coef(i1, j1) * i ^ j1 * i0 ^ (i1 + 1 - j1))' + countFun(j1 + 1, i1, λ j0. binom_coef(i1, j0 - 1) * i ^ j0 * i0 ^ (i1 + 1 - 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: 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