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Proof step 16 from proof countFun-mult (previous step)

i2 < 0, ∀ 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)) ⊦ 0 = Σ countFun(i, i2 + i, λ j0. i0 * intfun(j0))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: intlist inst: name: countFun-base seq: ⊦ j < i → countFun(i, j, intfun) = [] subst: [intfun, j, i] → [λ j0. i0 * intfun(j0), i2 + i, i] flags: rewrite, current

Additional information

Used simplifier rules: sum-empty: ⊦ Σ [] = 0;%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: λ j0. i0 * intfun(j0), i2 + i, i