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Proof step 1 from proof countFun-length (conclusion)

⊦ 0 ≤ i ∧ i ≤ j → # countFun(i, j, intfun) = i→n(1 + j - i)

User applied rule induction

And obtained the following new goals

Proof step 2:
true, ∀ j0, i0, intfun. i→n(j0) - i→n(i0) < i→n(j) - i→n(i) → true → 0 ≤ i0 ∧ i0 ≤ j0 → # countFun(i0, j0, intfun) = i→n(1 + j0 - i0) ⊦ 0 ≤ i ∧ i ≤ j → # countFun(i, j, intfun) = i→n(1 + j - i)

Rule argument

indvar: i→n(j) - i→n(i) wfpred: <

Additional information

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