[Imprint][Data Privacy Statement][E-Mail]

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Proof step 5 from proof sum-constant-countFun (previous step)

i ≤ j, 0 ≤ 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

User applied rule apply induction

And obtained the following new goals

Proof step 6:
i ≤ j, 0 ≤ 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 ⊦ i→n(j - 1) - i→n(i) < i→n(j) - i→n(i), Σ countFun(i, j - 1, intfun) + intfun(j) = (1 + j - i) * i0
Proof step 18:
i ≤ j, 0 ≤ 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 ⊦ (∀ i2. i ≤ i2 ∧ i2 ≤ j - 1 → intfun(i2) = i0) ∧ i ≤ j - 1 ∧ 0 ≤ i, Σ countFun(i, j - 1, intfun) + intfun(j) = (1 + j - i) * i0
Proof step 22:
Σ countFun(i, j - 1, intfun) = (1 + (j - 1) - i) * i0, i ≤ j, 0 ≤ 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

Rule argument

subst: [j0, i1, intfun, i0] → [j - 1, i, intfun, i0]

Additional information

Used terms: j - 1, i, intfun, i0