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Proof step 1 from proof countFun-last-01 (conclusion)

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

User applied rule induction

And obtained the following new goals

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

Rule argument

indvar: i→n(j - i) wfpred: <

Additional information

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