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Proof step 1 from proof countFun-dist (conclusion)

⊦ 0 ≤ i ∧ i ≤ j → Σ countFun(i, j, λ j0. intfun1(j0) + intfun2(j0)) = Σ countFun(i, j, λ j0. intfun1(j0)) + Σ countFun(i, j, λ j0. intfun2(j0))

User applied rule induction

And obtained the following new goals

Proof step 2:
true, ∀ j0, i0, intfun1, intfun2. i→n(j0 - i0) < i→n(j - i) → 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)) ⊦ 0 ≤ i ∧ i ≤ j → Σ countFun(i, j, λ j0. intfun1(j0) + intfun2(j0)) = Σ countFun(i, j, λ j0. intfun1(j0)) + Σ countFun(i, j, λ j0. intfun2(j0))

Rule argument

indvar: i→n(j - i) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j - i)