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Proof step 17 from proof countFun-dist (previous step)

0 ≤ i, 0 ≤ i1 + i, 0 ≤ i1, ∀ j0, i0, intfun1, intfun2. i→n(j0 - i0) < i→n(i1) → true → 0 ≤ i0 ∧ i0 ≤ j0 → Σ countFun(i0, j0, λ j0. intfun1(j0) + intfun2(j0)) = Σ countFun(i0, j0, λ j0. intfun1(j0)) + Σ countFun(i0, j0, λ j0. intfun2(j0)) ⊦ Σ countFun(i + 1, i1 + i, λ j0. intfun1(j0) + intfun2(j0)) = Σ countFun(i + 1, i1 + i, intfun1) + Σ countFun(i + 1, i1 + i, intfun2)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [intfun2, intfun1, i0, j0] → [intfun2, intfun1, i + 1, i1 + i]

Additional information

Used simplifier rules: countFun-base-01: ⊦ countFun(i + 1, i, intfun) = [];%sum-empty: ⊦ Σ [] = 0;%countFun-base-01: ⊦ countFun(i + 1, i, intfun) = [];%sum-empty: ⊦ Σ [] = 0;%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; 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%le-add: ⊦ i + j ≤ i0 + j ↔ i ≤ i0; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%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%sub-add-05: ⊦ i + j - i + 1 = j - 1; from specification int-abs%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat Used terms: intfun2, intfun1, i + 1, i1 + i