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Proof step 6 from proof countFun-at (previous step)

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))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ n = # ints → (ints + i ')[n] = i
Proof step 8:
n ≠ # countFun(i, j - 1, intfun), 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))
Proof step 37:
n = # countFun(i, j - 1, intfun), 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)) ⊦ intfun(j) = intfun(i + n→i(n))

Rule argument

spec: list-del inst: intlista name: get-last seq: ⊦ n = # ints → (ints + i ')[n] = i subst: [n, i, ints] → [n, intfun(j), countFun(i, j - 1, intfun)] flags: rewrite

Additional information

Used terms: n, intfun(j), countFun(i, j - 1, intfun)