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Proof step 34 from proof countFun-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: l: ⊦ # ints = 0 ↔ ints = []; from specification list-data[intlista]%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%ler: ⊦ m ≤ m; from specification nat%le: ⊦ n→i(m) ≤ n→i(n) ↔ m ≤ n; from specification intnat%le-add: ⊦ i + j ≤ i0 + j ↔ i ≤ i0; from specification int-abs%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n; from specification intnat%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%nat2int-02: ⊦ n→i(n) = 0 ↔ n = 0; from specification intnat%eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs