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

i ≤ i1 + i, 0 ≤ i, 0 ≤ i1 + i, 0 ≤ i1, ∀ j0, i0, intfun1, intfun2. i→n(j0 - i0) < i→n(i1) → true → 0 ≤ i0 ∧ i0 ≤ j0 → Σ countFun(i0, j0, λ j0. intfun1(j0) + intfun2(j0)) = Σ countFun(i0, j0, λ j0. intfun1(j0)) + Σ countFun(i0, j0, λ j0. intfun2(j0)) ⊦ Σ((intfun1(i) + intfun2(i))' + countFun(i + 1, i1 + i, λ j0. intfun1(j0) + intfun2(j0))) = Σ countFun(i, i1 + i, intfun1) + Σ countFun(i, i1 + i, intfun2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ i, 0 ≤ i1 + i, 0 ≤ i1, ∀ j0, i0, intfun1, intfun2. i→n(j0 - i0) < i→n(i1) → true → 0 ≤ i0 ∧ i0 ≤ j0 → Σ countFun(i0, j0, λ j0. intfun1(j0) + intfun2(j0)) = Σ countFun(i0, j0, λ j0. intfun1(j0)) + Σ countFun(i0, j0, λ j0. intfun2(j0)) ⊦ (intfun1(i) + intfun2(i)) + Σ countFun(i + 1, i1 + i, λ j0. intfun1(j0) + intfun2(j0)) = Σ countFun(i, i1 + i, intfun1) + Σ countFun(i, i1 + i, intfun2)

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]