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Proof step 2 from proof sum-countFun-sub (previous step)

0 ≤ i ⊦ Σ countFun(i, j, intfun) = Σ countFun(i, j, λ i0. intfun(i0) - intfun0(i0)) + Σ countFun(i, j, intfun0)

User applied rule induction

And obtained the following new goals

Proof step 3:
0 ≤ i, ∀ j0, i, intfun, intfun0. i→n(j0) < i→n(j) → 0 ≤ i → Σ countFun(i, j0, intfun) = Σ countFun(i, j0, λ i0. intfun(i0) - intfun0(i0)) + Σ countFun(i, j0, intfun0) ⊦ Σ countFun(i, j, intfun) = Σ countFun(i, j, λ i0. intfun(i0) - intfun0(i0)) + Σ countFun(i, j, intfun0)

Rule argument

indvar: i→n(j) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j)