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Proof step 23 from proof rightshift-8-01-01 (previous step)

¬(0 ≤ i ∧ int2bytes(i).first = i2b(0)), 0 ≤ i, # int2bytes(i) ≤ i→n(j), 2 ^ (8 * abs(j)) ≤ i ⊦ j = 0, i / 2 ^ (8 * abs(j)) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
0 ≤ i, # int2bytes(i) ≤ i→n(j), 2 ^ (8 * abs(j)) ≤ i ⊦ j = 0, i / 2 ^ (8 * abs(j)) = 0, int2bytes(i).first = i2b(0)

Rule argument

Additional information

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