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Proof step 2 from proof shiftright-pos-div (previous step)

0 ≤ i, 0 ≤ j ⊦ i >> j = i / 2 ^ j

User applied rule induction

And obtained the following new goals

Proof step 3:
0 ≤ i ∧ 0 ≤ j, ∀ j, i0. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j → i0 >> j = i0 / 2 ^ j ⊦ i >> j = i / 2 ^ j

Rule argument

indvar: i→n(i) wfpred: <

Additional information

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