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Proof step 28 from proof countFun-at (previous step)

0 < j ∧ j ≠ 1, i→n(j) -1 = n0 + i→n(i), i ≤ abs(j - 1), i→n(j) ≤ n0 + i→n(i), n < # countFun(i, j - 1, intfun), i ≤ j, n→i(n) + i ≤ j, 0 ≤ i, 0 ≤ 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 = # countFun(i, j - 1, intfun), countFun(i, j - 1, intfun)[n] = intfun(i + n→i(n))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) 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%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%abs-09: ⊦ 0 < i → abs(i - 1) = abs(i) - 1; from specification int-abs%pred-03: ⊦ n -1 = n - 1; from specification nat