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Proof step 6 from proof shiftright-pos-div (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 ≤ j, 0 ≤ i, 0 < j, ∀ j, i0. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j → i0 >> j = i0 / 2 ^ j ⊦ i / 2 >> (j - 1) = i / 2 ^ j

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)