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Proof step 17 from proof twopot-mone-div (previous step)

0 < 2 ^ (i - 1) ∧ 0 < 2 ^ (j - 1), (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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
(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

Rule argument

Additional information

Used simplifier rules: pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs