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Proof step 15 from proof countFun-at (previous step)

0 ≤ j, 0 ≤ i, n→i(n) + i ≤ j, i ≤ j, i < j, i→n(j - i) < n, 0 < 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) '[n - i→n(j - i)] = intfun(i + n→i(n)), n = i→n(j - i)

User applied rule cut formula

And obtained the following new goals

Proof step 16:
0 ≤ j, 0 ≤ i, n→i(n) + i ≤ j, i ≤ j, i < j, i→n(j - i) < n, 0 < 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)) ⊦ n→i(n) ≤ j - i, intfun(j) '[n - i→n(j - i)] = intfun(i + n→i(n)), n = i→n(j - i)

Rule argument

fma: n→i(n) ≤ j - i

Additional information

Used simplifier rules: inni-ls: ⊦ i→n(i) < n → ¬ n→i(n) ≤ i; from specification intnat Used formulas: n→i(n) ≤ j - i