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Proof step 19 from proof twopot-mone-div (previous step)

(2 * 2 ^ (i - 1)) / (2 * 2 ^ (j - 1)) = 2 ^ (i - 1) / 2 ^ (j - 1), 0 < j, 0 < i, j ≤ i, 0 ≤ j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → 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: ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i - j ≤ i0 ↔ i ≤ j + i0; (s)% ⊦ i + (j - i) = j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; (s)%j ≠ 0, i0 ≠ 0 ⊦ (i * i0) / (j * i0) = i / j; (s)%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