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Proof step 24 from proof sum-constant-countFun (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
Σ countFun(i, j - 1, intfun) = (j - i) * intfun(j), 0 ≤ j, 0 ≤ i, i ≤ j, ∀ i1. i ≤ i1 ∧ i1 ≤ j → intfun(i1) = intfun(j), ∀ 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 ⊦ (j - i) * intfun(j) + intfun(j) = (1 + j - i) * intfun(j)

Rule argument

Additional information

Used simplifier rules: ler: ⊦ i ≤ i; from specification int-abs