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Proof step 31 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)) ⊦ 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 ∧ 0 < j

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: j = 0

Additional information

Used simplifier rules: countFun-base-03: ⊦ countFun(0, ~1, intfun) = [];%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%size-01: ⊦ # [] = 0; from specification list-data[intlista] Used formulas: j = 0