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Proof step 7 from proof shiftright-byte-8-pos-01 (previous step)

¬ abs(b2i(by)) < abs(2 ^ j), 0 ≤ j, 0 < j, 0 ≤ b2i(by), 8 ≤ j ⊦ b2i(by) / 2 ^ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 ≤ j, 0 < j, 0 ≤ b2i(by), 8 ≤ j, 2 ^ j ≤ b2i(by) ⊦ b2i(by) / 2 ^ j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ abs(2 ^ i) = 2 ^ i;% ⊦ 0 ≤ i → abs(i) = i;