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Proof step 30 from proof countFun-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ 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%succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs