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Proof step 23 from proof countFun-at (previous step)

0 ≤ j, 0 ≤ i, n→i(n) + i ≤ j, i ≤ j, n < # countFun(i, j - 1, intfun), i→n(j) ≤ (i→n(j - 1) - i→n(i)) + i→n(i), abs(j - 1) < i, abs(j - 1) < 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)[n] = intfun(i + n→i(n)), n = # countFun(i, j - 1, intfun), countFun(i, j - 1, intfun) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [j - 1] flags: rewrite

Additional information

Used simplifier rules: countFun-base-02: ⊦ countFun(j, j - 1, intfun) = [];%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ 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%ler: ⊦ i ≤ i; from specification int-abs%bet-01: ⊦ 0 ≤ i → (i < 1 ↔ i = 0); from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs Used terms: j - 1