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

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 11 from proof sum-constant-countFun (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
0 ≤ i, i ≤ 1, 1 ≤ (0 - i→n(i)) + i→n(i), ∀ i1. i ≤ i1 ∧ i1 ≤ 1 → intfun(i1) = i0, ∀ j0, i1, intfun, i0. i→n(j0) - i→n(i1) < i→n(1) - 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, 0, intfun) + intfun(1) = (2 - i) * i0, (0 - i→n(i)) + i→n(i) = 0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2