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Proof step 1 from proof sum-constant-countFun (conclusion)

∀ i1. i ≤ i1 ∧ i1 ≤ j → intfun(i1) = i0, i ≤ j, 0 ≤ i ⊦ Σ countFun(i, j, intfun) = (1 + j - i) * i0

User applied rule induction

And obtained the following new goals

Proof step 2:
(∀ i1. i ≤ i1 ∧ i1 ≤ j → intfun(i1) = i0) ∧ i ≤ j ∧ 0 ≤ i, ∀ j0, i1, intfun, i0. i→n(j0) - i→n(i1) < i→n(j) - 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, j, intfun) = (1 + j - i) * i0

Rule argument

indvar: i→n(j) - i→n(i) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j) - i→n(i)