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Proof step 1 from proof twopot-mone-div (conclusion)

⊦ j ≤ i ∧ 0 ≤ j → (2 ^ i - 1) / 2 ^ j = 2 ^ i / 2 ^ j - 1

User applied rule induction

And obtained the following new goals

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

Rule argument

indvar: i→n(i) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(i)