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Proof step 20 from proof sum-countFun-sub (previous step)

i ≤ i1 + 1, ~1 ≤ i1, 0 ≤ i, i ≤ i1 + 1, ∀ j0, i, intfun, intfun0. i→n(j0) < i→n(i1 + 1) → 0 ≤ i → Σ countFun(i, j0, intfun) = Σ countFun(i, j0, λ i0. intfun(i0) - intfun0(i0)) + Σ countFun(i, j0, intfun0) ⊦ Σ countFun(i, i1, intfun) + intfun(i1 + 1) = Σ(countFun(i, i1 + 1 - 1, λ i0. intfun(i0) - intfun0(i0)) + (intfun(i1 + 1) - intfun0(i1 + 1))') + Σ countFun(i, i1 + 1, intfun0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sum-app: ⊦ Σ(ints + ints0) = Σ ints + Σ ints0;%sum-one: ⊦ Σ i ' = i;%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ax-04: ⊦ i + 0 = i; from specification int-basic2%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs