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Proof step 8 from proof byte2bits-val (
previous step
)
¬(# int2bits(b2i(by)) ≤ 8 ∧ int2bits(b2i(by)) ≠ []), b2i(by) < 0 ⊦ bits2uint(fillsgn(8, int2bits(b2i(by)))) = b2i(by) + 256
Heuristic simplifier applied rule simplifier
And closed the goal!
Rule argument
Additional information
Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ ¬ 8 < # int2bits(b2i(by));% ⊦ ¬ m ≤ n ↔ n < m;