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Proof step 8 from proof countFun-length (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: intnat inst: name: Int2nat-pos seq: ⊦ 0 < i → i→n(i) = i→n(i - 1) + 1 subst: [i] → [1 + j - i] flags: rewrite

Additional information

Used simplifier rules: countFun-base-01: ⊦ countFun(i + 1, i, intfun) = [];%sub-sub: ⊦ i - j - i0 = i - j + i0; from specification int-abs%sub-add-06: ⊦ i + j - i0 + j = i - i0; 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%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%sub-eq-01: ⊦ i = j → i - j = 0; from specification int-abs%l: ⊦ # ints = 0 ↔ ints = []; from specification list-data[intlista]%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs Used terms: 1 + j - i