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Proof step 4 from proof countFun-mult (previous step)

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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ 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, i2 + i, intfun) = Σ countFun(i, i2 + i, λ j0. i0 * intfun(j0))

Proof step 4 from proof countFun-mult (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information