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Proof step 14 from proof sum-constant-countFun (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i ≤ j, 0 ≤ i, 0 ≤ j, i→n(j) ≤ (i→n(j) - 1 - i→n(i)) + i→n(i), 0 < j, ∀ 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 ⊦ Σ countFun(i, j - 1, intfun) + intfun(j) = (1 + j - i) * i0, j = 1

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%pred-03: ⊦ n -1 = n - 1; from specification nat