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Proof step 1 from proof shiftright-byte-8-pos-01 (conclusion)

⊦ 0 ≤ b2i(by) ∧ 8 ≤ j → b2i(by) >> j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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