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Proof step 26 from proof rightshift-8-01-01 (previous step)

# int2bytes(i) < i→n(j) + # int2bytes(i) ∧ bytes2int(int2bytes(i)) < 0, # int2bytes(i) ≤ i→n(j) ⊦ i >> 8 * j = bytes2int(firstn(# int2bytes(i), mklist(i2b(~1), i→n(j) + # int2bytes(i) - # int2bytes(i)) + int2bytes(i)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: int2int: ⊦ bytes2int(int2bytes(i)) = i;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m < m + n ↔ n ≠ 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;