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Proof step 34 from proof rightshift-8-01 (previous step)

i→n(j) = 1, i < ~128, ~256 < i, ~256 ≤ i, i < 0, ∀ i, j0. i→n(j0) < 1 → i→n(j0) < # int2bytes(i) → i >> 8 * abs(j0) = bytes2int(firstn(# int2bytes(i) - i→n(j0), int2bytes(i))) ⊦ abs(j) = 1, (i / 256 - 1) >> 8 * (abs(j) - 1) = ~1, j = 0, i % 256 = 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] → [1, j] rotate: F spec: intnat inst: name: num-i2n-eq

Additional information

Used simplifier rules: ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j → (i ≠ j ↔ j < i);% ⊦ ¬ j < 0 → abs(j) = j;%i < 0 ⊦ i→n(i) = (* m) ↔ ~ i = n→i(m);% ⊦ ~ i = j ↔ i + j = 0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i < 0 → abs(i) = ~ i; Used terms: 1, j