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Proof step 25 from proof countFun-at (previous step)

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

Proof step 26:
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))

Rule argument

Additional information

Used simplifier rules: abs-03: ⊦ 0 ≤ abs(i); from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%labs-01: ⊦ i→n(i) ≤ i→n(j) ↔ abs(i) ≤ abs(j); from specification intnat