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Proof step 113 from proof rightshift-8-01 (previous step)

# int2bytes(i / 256) ≤ i→n(abs(j) - 1), 0 ≤ i, 255 < i, i→n(j) ≤ # int2bytes(i / 256), 0 < i, abs(j) ≤ abs(abs(j) - 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))) ⊦ j = 0, i / 256 >> 8 * (abs(j) - 1) = bytes2int(firstn((# int2bytes(i / 256) - i→n(j)) + 1, int2bytes(i / 256) + i2b(i % 256) '))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 114:
j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j)

Rule argument

seq: j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j) subst: [j, i] → [1, abs(j)] rotate: F spec: intnat inst: name: sub

Additional information

Used simplifier rules: ⊦ 1 ≤ i ↔ 0 < i;% ⊦ 0 < abs(i) ↔ i ≠ 0;% ⊦ i→n(abs(i)) = i→n(i);% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m < 1 ↔ m = 0;% ⊦ ¬ m ≤ n ↔ n < m;%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n;% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ m ≤ n → (n < m ↔ false);% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: 1, abs(j)