[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 18 from proof twopot-mone-div (previous step)

(2 * 2 ^ (i - 1)) / (2 * 2 ^ (j - 1)) = 2 ^ (i - 1) / 2 ^ (j - 1), j ≤ i, 0 ≤ i, 0 ≤ j, 0 < i, 0 < j, ∀ i0, j. i→n(i0) < i→n(i) → true → j ≤ i0 ∧ 0 ≤ j → (2 ^ i0 - 1) / 2 ^ j = 2 ^ i0 / 2 ^ j - 1 ⊦ 2 ^ (i - 1) / 2 ^ (j - 1) = (2 ^ (i - 1) - 1) / 2 ^ (j - 1) + 1

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i0, j] → [i - 1, j - 1]

Additional information

Used simplifier rules: sub-le: ⊦ i - j ≤ i0 ↔ i ≤ j + i0; from specification int-abs%Sub-def: ⊦ i + (j - 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%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%div-mult-01: j ≠ 0, i0 ≠ 0 ⊦ (i * i0) / (j * i0) = i / j; from specification int-div%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: i - 1, j - 1