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Proof step 16 from proof sum-countFun-sub (previous step)

i ≤ j, 0 ≤ i, 0 ≤ j, ∀ 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 countFun(i, j - 1, intfun) + intfun(j) = Σ countFun(i, j, λ i0. intfun(i0) - intfun0(i0)) + Σ countFun(i, j, intfun0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, 1, i1] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 1, i1