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Proof step 7 from proof countFun-length (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sub-add-06: ⊦ i + j - i0 + j = i - i0; from specification int-abs%succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs