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Proof step 40 from proof rightshift-8-01-01 (previous step)

¬(i < 0 ∧ int2bytes(i).first ≠ i2b(~1)), i < 0, # int2bytes(i) ≤ i→n(j), 2 ^ (8 * abs(j) - 1) + i < 0 ⊦ j = 0, i >> 8 * j = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
int2bytes(i).first = i2b(~1), i < 0, # int2bytes(i) ≤ i→n(j), 2 ^ (8 * abs(j) - 1) + i < 0 ⊦ j = 0, i >> 8 * j = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;