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Proof step 11 from proof countFun-mult (previous step)

0 ≤ i2, ∀ 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 * (intfun(i) + Σ countFun(i + 1, i2 + i, intfun)) = 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-rec seq: ⊦ i ≤ j → countFun(i, j, intfun) = intfun(i) ' + countFun(i + 1, j, intfun) subst: [intfun, j, i] → [λ j0. i0 * intfun(j0), i2 + i, i] flags: rewrite, current

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lenls: ⊦ i ≤ j → ¬ j < i; 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