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Proof step 3 from proof twopot-mone-div (previous step)

j ≤ i, 0 ≤ j, ∀ 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 = 2 ^ i / 2 ^ j - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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 / 2 ^ j = (2 ^ i - 1) / 2 ^ j + 1

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i - j = i0 ↔ i = i0 + j; (s)