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Proof step 10 from proof sum-constant-countFun (previous step)

i→n(j) ≤ (i→n(j - 1) - i→n(i)) + i→n(i), 0 ≤ j, 0 ≤ i, i ≤ 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, 0 < j ∧ j ≠ 1

User applied rule cut formula

And obtained the following new goals

Proof step 11:
j = 1, i→n(j) ≤ (i→n(j - 1) - i→n(i)) + i→n(i), 0 ≤ j, 0 ≤ i, i ≤ 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, 0 < j ∧ j ≠ 1

Rule argument

fma: j = 1

Additional information

Used simplifier rules: countFun-base-03: ⊦ countFun(0, ~1, intfun) = [];%sum-empty: ⊦ Σ [] = 0;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lez: ⊦ 0 ≤ m; from specification nat%ax-04: ⊦ i + 0 = i; from specification int-basic2%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 Used formulas: j = 1