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Proof step 9 from proof sum-countFun-sub (previous step)

j < i, 0 ≤ i, j < i, ∀ j0, i, intfun, intfun0. i→n(j0) < i→n(j) → 0 ≤ i → Σ countFun(i, j0, intfun) = Σ countFun(i, j0, λ i0. intfun(i0) - intfun0(i0)) + Σ countFun(i, j0, intfun0) ⊦ Σ [] = Σ countFun(i, j, λ i0. intfun(i0) - intfun0(i0)) + Σ countFun(i, j, intfun0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sum-empty: ⊦ Σ [] = 0;