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Proof step 12 from proof countFun-at (previous step)

# countFun(i, j - 1, intfun) < n, 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) '[n - # countFun(i, j - 1, intfun)] = intfun(i + n→i(n)), n = # countFun(i, j - 1, intfun)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: countFun-length seq: ⊦ 0 ≤ i ∧ i ≤ j → # countFun(i, j, intfun) = i→n(1 + j - i) subst: [intfun, j, i] → [intfun, j - 1, i] flags: rewrite, current

Additional information

Used simplifier rules: countFun-base-02: ⊦ countFun(j, j - 1, intfun) = [];%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%size-01: ⊦ # [] = 0; from specification list-data[intlista]%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%ax-10: ⊦ m - 0 = m; from specification nat%ler: ⊦ i ≤ i; from specification int-abs%lep-01: ⊦ i + j ≤ i ↔ j ≤ 0; from specification int-abs%nat2int-10: ⊦ n→i(n) ≤ 0 ↔ n = 0; from specification intnat Used terms: intfun, j - 1, i