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Proof step 1 from proof countFun-mult (conclusion)

⊦ i0 * Σ countFun(i, j, intfun) = Σ countFun(i, j, λ j0. i0 * intfun(j0))

User applied rule induction

And obtained the following new goals

Proof step 2:
true, ∀ 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))

Rule argument

indvar: i→n(j - i) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j - i)