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Proof step 14 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)) ⊦ Σ countFun(i + 1, i2 + i, intfun) * i0 = Σ countFun(i + 1, i2 + i, λ j0. i0 * intfun(j0))

User applied rule apply induction

And closed the goal!

Rule argument

subst: [intfun, i0, i1, j0] → [intfun, i0, i + 1, i2 + i]

Additional information

Used simplifier rules: countFun-base-01: ⊦ countFun(i + 1, i, intfun) = [];%sum-empty: ⊦ Σ [] = 0;%commutativity: ⊦ i * j = j * i; from specification int-mult%sub-add-05: ⊦ i + j - i + 1 = j - 1; 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%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2%Mult-zero: ⊦ i * 0 = 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult Used terms: intfun, i0, i + 1, i2 + i