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Proof step 4 from proof countFun-at (previous step)

0 ≤ j, 0 ≤ i, n→i(n) + i ≤ j, i ≤ j, ∀ j0, i0, n, intfun. i→n(j0) - i→n(i0) < i→n(j) - i→n(i) → i0 ≤ j0 ∧ n→i(n) + i0 ≤ j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 → countFun(i0, j0, intfun)[n] = intfun(i0 + n→i(n)) ⊦ countFun(i, j, intfun)[n] = intfun(i + n→i(n))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ i ≤ j → countFun(i, j, intfun) = countFun(i, j - 1, intfun) + intfun(j) '
Proof step 6:
i ≤ j, 0 ≤ j, 0 ≤ i, n→i(n) + i ≤ j, i ≤ j, ∀ j0, i0, n, intfun. i→n(j0) - i→n(i0) < i→n(j) - i→n(i) → i0 ≤ j0 ∧ n→i(n) + i0 ≤ j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 → countFun(i0, j0, intfun)[n] = intfun(i0 + n→i(n)) ⊦ (countFun(i, j - 1, intfun) + intfun(j) ')[n] = intfun(i + n→i(n))

Rule argument

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

Additional information

Used terms: intfun, j, i