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Proof step 76 from proof rightshift-8-01-01 (previous step)

¬(# int2bytes(i) < i→n(j) + # int2bytes(i) ∧ bytes2int(int2bytes(i)) < 0), 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)))), j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: int2int: ⊦ bytes2int(int2bytes(i)) = i;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m + m0 ≤ m ↔ m0 = 0;% ⊦ ¬ m < n ↔ n ≤ m;