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Proof step 35 from proof countFun-at (previous step)

0 < n→i(n) + i, n < # countFun(i, n→i(n) + i - 1, intfun), 0 ≤ i, ∀ j0, i0, n0, intfun. i→n(j0) - i→n(i0) < i→n(n→i(n) + i) - i→n(i) → i0 ≤ j0 ∧ n→i(n0) + i0 ≤ j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 → countFun(i0, j0, intfun)[n0] = intfun(i0 + n→i(n0)) ⊦ n = 0, n→i(n) + i = 0, countFun(i, n→i(n) + i - 1, intfun)[n] = intfun(i + n→i(n)), n = # countFun(i, n→i(n) + i - 1, intfun), countFun(i, n→i(n) + i - 1, intfun) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: countFun-length seq: ⊦ 0 ≤ i ∧ i ≤ j → # countFun(i, j, intfun) = i→n(1 + j - i) subst: [intfun, j, i] → [intfun, n→i(n) + i - 1, i] flags: rewrite, current

Additional information

Used simplifier rules: Sub-def: ⊦ i + (j - i) = j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-add: ⊦ i + j - i = j; from specification int-abs%n2n: ⊦ i→n(n→i(n)) = n; from specification intnat%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%le-add: ⊦ i + j ≤ i0 + j ↔ i ≤ i0; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n; from specification intnat%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1 Used terms: intfun, n→i(n) + i - 1, i