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Proof step 6 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), 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: int2int: ⊦ bytes2int(int2bytes(i)) = i;%int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m ≤ 0 ↔ m = 0;% ⊦ 0 + m = m;% ⊦ i >> 0 = i;% ⊦ ¬ 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;