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Proof step 3 from proof countFun-length (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i ≤ j → countFun(i, j, intfun) = intfun(i) ' + countFun(i + 1, j, intfun)
Proof step 5:
i ≤ j, i ≤ j, 0 ≤ i, 0 ≤ j, ∀ 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) ⊦ #(intfun(i) ' + countFun(i + 1, j, intfun)) = i→n(1 + j - i)

Rule argument

spec: inst: name: countFun-rec seq: ⊦ i ≤ j → countFun(i, j, intfun) = intfun(i) ' + countFun(i + 1, j, intfun) subst: [intfun, j, i] → [intfun, j, i] flags: rewrite, current

Additional information

Used terms: intfun, j, i