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Proof step 21 from proof sum-countFun-sub (previous step)

~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 + 1, intfun0) + Σ countFun(i, i1, λ i0. intfun(i0) - intfun0(i0)) = Σ countFun(i, i1, intfun) + intfun0(i1 + 1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ i ≤ j → countFun(i, j, intfun) = countFun(i, j - 1, intfun) + intfun(j) '
Proof step 23:
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 + 1 - 1, intfun0) + intfun0(i1 + 1) ') + Σ countFun(i, i1, λ i0. intfun(i0) - intfun0(i0)) = Σ countFun(i, i1, intfun) + intfun0(i1 + 1)

Rule argument

spec: inst: name: countFun-last seq: ⊦ i ≤ j → countFun(i, j, intfun) = countFun(i, j - 1, intfun) + intfun(j) ' subst: [intfun, j, i] → [intfun0, i1 + 1, i] flags: rewrite, current

Additional information

Used terms: intfun0, i1 + 1, i