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Proof step 10 from proof countFun-mult (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ i2, ∀ j0, i1, i0, intfun. i→n(j0 - i1) < i→n(i2) → true → i0 * Σ countFun(i1, j0, intfun) = Σ countFun(i1, j0, λ j0. i0 * intfun(j0)) ⊦ i0 * (intfun(i) + Σ countFun(i + 1, i2 + i, intfun)) = Σ countFun(i, i2 + i, λ j0. i0 * intfun(j0))

Rule argument

Additional information

Used simplifier rules: sum-rec: ⊦ Σ(i ' + ints) = i + Σ ints;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]