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Proof step 7 from proof countFun-dist (previous step)

¬ i1 + i < 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 countFun(i, 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

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%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2