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Proof step 12 from proof sum-constant-countFun (previous step)

0 ≤ i, i ≤ 1, 1 ≤ (0 - i→n(i)) + i→n(i), ∀ i1. i ≤ i1 ∧ i1 ≤ 1 → intfun(i1) = i0, ∀ j0, i1, intfun, i0. i→n(j0) - i→n(i1) < i→n(1) - i→n(i) → (∀ i2. i1 ≤ i2 ∧ i2 ≤ j0 → intfun(i2) = i0) ∧ i1 ≤ j0 ∧ 0 ≤ i1 → Σ countFun(i1, j0, intfun) = (1 + j0 - i1) * i0 ⊦ Σ countFun(i, 0, intfun) + intfun(1) = (2 - i) * i0, (0 - i→n(i)) + i→n(i) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: countFun-base seq: ⊦ j < i → countFun(i, j, intfun) = [] subst: [intfun, j, i] → [intfun, 0, i] flags: rewrite, current

Additional information

Used simplifier rules: sum-empty: ⊦ Σ [] = 0;%ax-04: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%num-i2n-le: 0 ≤ i ⊦ i→n(i) ≤ (* m) ↔ i ≤ n→i(m); from specification intnat%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%eq-02: n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; from specification nat Used terms: intfun, 0, i