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Proof step 15 from proof countFun-mult (previous step)

i2 + i < 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 [] = Σ countFun(i, i2 + i, λ j0. i0 * intfun(j0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sum-empty: ⊦ Σ [] = 0;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%Mult-zero: ⊦ i * 0 = 0; from specification int-mult