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Proof step 10 from proof count-i2j-length (previous step)

# count-i2j(i + 1, j) = i→n(j - i + 1) + 1, i ≤ j, ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → i0 ≤ j0 → # count-i2j(i0, j0) = i→n(j0 - i0) + 1 ⊦ # count-i2j(i + 1, j) = i→n(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] → [j - i] flags: rewrite

Additional information

Used simplifier rules: count-i2j-empty-02: ⊦ count-i2j(i, j) = [] ↔ j < i;%sub-sub: ⊦ i - j - i0 = i - j + i0; from specification int-abs%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%add-zero: ⊦ i + 0 = i; from specification int-basic2%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%sub-eq-01: ⊦ i = j → i - j = 0; from specification int-abs%l: ⊦ # ints = 0 ↔ ints = []; from specification list-data[intlista]%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs Used terms: j - i