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Proof step 3 from proof countFun-mult (previous step)

∀ j0, i1, i0, intfun. i→n(j0 - i1) < i→n(j - i) → true → i0 * Σ countFun(i1, j0, intfun) = Σ countFun(i1, j0, λ j0. i0 * intfun(j0)) ⊦ i0 * Σ countFun(i, j, intfun) = Σ countFun(i, j, λ j0. i0 * intfun(j0))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
j = i2 + i, ∀ j0, i1, i0, intfun. i→n(j0 - i1) < i→n(i2) → true → i0 * Σ countFun(i1, j0, intfun) = Σ countFun(i1, j0, λ j0. i0 * intfun(j0)) ⊦ i0 * Σ countFun(i, j, intfun) = Σ countFun(i, j, λ j0. i0 * intfun(j0))

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, i, i2] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, i, i2