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Proof step 11 from proof int2bytes-butlast (previous step)

¬ i ≤ ~256 ∧ ¬ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ (i2b(~1) ' + i2b(i) ').butlast = int2bytes(i >> 8)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
i < ~128, ~256 < i, i < 0 ⊦ i >> 8 = ~1

Rule argument

Additional information

Used simplifier rules: int2bytes-minus-one: ⊦ int2bytes(i) = i2b(~1) ' ↔ i = ~1;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)%j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ (bytes + by ').butlast = bytes;