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Proof step 73 from proof rightshift-8-01-01 (previous step)

¬ # int2bytes(i) ≥ i→n(j) + # int2bytes(i), i→n(j) < # int2bytes(i) ⊦ bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i))) = bytes2int(firstn(# int2bytes(i), fillsgn(i→n(j) + # int2bytes(i), int2bytes(i))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m + m0 ≤ m ↔ m0 = 0;% ⊦ n ≥ m ↔ m ≤ n;