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Proof step 2 from proof countFun-dist (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i ≤ j, 0 ≤ i, 0 ≤ j, ∀ j0, i0, intfun1, intfun2. i→n(j0 - i0) < i→n(j - i) → 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, j, λ j0. intfun1(j0) + intfun2(j0)) = Σ countFun(i, j, intfun1) + Σ countFun(i, j, intfun2)

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs