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Proof step 105 from proof rightshift-8-01 (previous step)

0 < i, i→n(j) < 2, 127 < i, i ≤ 255, 0 ≤ i, ∀ 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 = 1, i / 256 >> 8 * (abs(j) - 1) = bytes2int(firstn(2 - i→n(j), i2b(0) ' + i2b(i - 256) ')), j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: 0 ≤ i ⊦ i→n(i) < (* m) ↔ i < n→i(m) subst: [m, i] → [2, j] rotate: F spec: intnat inst: name: num-i2n-ls

Additional information

Used simplifier rules: bytes2int-zero: ⊦ bytes2int(i2b(0) ') = 0;% ⊦ 0 < j → (j < 2 ↔ j = 1);% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 0 ≤ i → abs(i) = i;% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ j < 0 → abs(j) = j;% ⊦ ¬ i < j → (i ≠ j ↔ j < i);%i < 0 ⊦ i→n(i) < (* m) ↔ ~ i < n→i(m);% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ ~ i - j = ~(i + j);% ⊦ ~ i * j = ~(i * j);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i < 0 → (~2 < i ↔ i = ~1);% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i >> 0 = i;% ⊦ firstn(1, by ' + bytes) = by ';% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ 0 < i ∧ 0 < j → (i / j = 0 ↔ i < j);%j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0); Used terms: 2, j