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Proof step 49 from proof rightshift-8-01 (previous step)

1 ≤ i→n(j) ∧ i→n(j) ≤ # int2bytes(i / 256 - 1), (i / 256 - 1) >> 8 * (abs(j) - 1) = bytes2int(firstn((# int2bytes(i / 256 - 1) - i→n(j)) + 1, int2bytes(i / 256 - 1))), i < 0, i < ~256, i→n(j) ≤ # int2bytes(i / 256 - 1), ∀ i, j0. i→n(j0) < i→n(j) → i→n(j0) < # int2bytes(i) → i >> 8 * abs(j0) = bytes2int(firstn(# int2bytes(i) - i→n(j0), int2bytes(i))) ⊦ i % 256 = 0, j = 0, (i / 256 - 1) >> 8 * (abs(j) - 1) = bytes2int(firstn((# int2bytes(i / 256 - 1) - i→n(j)) + 1, int2bytes(i / 256 - 1) + i2b(i % 256) '))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 50:
(i / 256 - 1) >> 8 * (abs(j) - 1) = bytes2int(firstn((# int2bytes(i / 256 - 1) - i→n(j)) + 1, int2bytes(i / 256 - 1))), 1 ≤ i→n(j), i < 0, i < ~256, i→n(j) ≤ # int2bytes(i / 256 - 1), 1 ≤ # int2bytes(i / 256 - 1), ∀ i, j0. i→n(j0) < i→n(j) → i→n(j0) < # int2bytes(i) → i >> 8 * abs(j0) = bytes2int(firstn(# int2bytes(i) - i→n(j0), int2bytes(i))) ⊦ i % 256 = 0, j = 0, (i / 256 - 1) >> 8 * (abs(j) - 1) = bytes2int(firstn((# int2bytes(i / 256 - 1) - i→n(j)) + 1, int2bytes(i / 256 - 1) + i2b(i % 256) '))

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ m + n = n + m;