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Proof step 10 from proof shiftright-byte-8-neg-01 (
previous step
)
abs(b2i(by)) < abs(256), 0 ≤ j, 0 < j, b2i(by) < 0, 8 ≤ j ⊦ (0 - 1) >> (j - 8) = ~1, b2i(by) = 0
Heuristic simplifier applied rule simplifier
And closed the goal!
Rule argument
Additional information
Used simplifier rules: ⊦ ~1 >> j = ~1;