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Proof step 20 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→n(j - 1) - i→n(i) < i→n(j) - i→n(i), 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: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls: n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; from specification nat%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; 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