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Proof step 7 from proof count-i2j-length (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
j ≤ abs(j - i + 1) + i, 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)

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%abs-03: ⊦ 0 ≤ abs(i); from specification int-abs%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%abssub: ⊦ j ≤ i → abs(i - j) = i - j; from specification int-abs%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat