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Proof step 6 from proof rightshift-8-02 (previous step)

¬(# int2bytes(i) < 2 ∧ bytes2int(int2bytes(i)) < 0), # int2bytes(i) = 1 ⊦ i >> 8 = bytes2int(fillsgn(2, int2bytes(i)).butlast)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# int2bytes(i) = 1, 0 ≤ i ⊦ i >> 8 = bytes2int(fillsgn(2, int2bytes(i)).butlast)

Rule argument

Additional information

Used simplifier rules: int2int: ⊦ bytes2int(int2bytes(i)) = i;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ m < n ↔ n ≤ m;