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Proof step 4 from proof shiftright-neg-div (previous step)

i + 2 ^ (abs(j) - 1) < 0 ⊦ j = 0, i >> abs(j) = (i + 1 - 2 ^ abs(j)) / 2 ^ abs(j)

User applied rule induction

And obtained the following new goals

Proof step 5:
i + 2 ^ (abs(j) - 1) < 0, ∀ i, j0. i→n(j0) < i→n(j) → i + 2 ^ (abs(j0) - 1) < 0 → j0 = 0 ∨ i >> abs(j0) = (i + 1 - 2 ^ abs(j0)) / 2 ^ abs(j0) ⊦ j = 0 ∨ i >> abs(j) = (i + 1 - 2 ^ abs(j)) / 2 ^ abs(j)

Rule argument

indvar: i→n(j) wfpred: <

Additional information

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