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Proof step 12 from proof int2bytes-first-zerobyte-01 (previous step)

# int2bits(j) = i→n(i0 * 8 + j0), j0 < 8, 0 ≤ j, 0 ≤ j0, 0 ≤ i0 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
0 ≤ i, 0 ≤ j ⊦ i→n(i + j) = i→n(i) + i→n(j)
Proof step 14:
0 ≤ i0 * 8 ∧ 0 ≤ j0, # int2bits(j) = i→n(i0 * 8) + i→n(j0), j0 < 8, 0 ≤ j, 0 ≤ j0, 0 ≤ i0 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

Rule argument

seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i + j) = i→n(i) + i→n(j) subst: [j, i] → [j0, i0 * 8] rotate: F spec: intnat inst: name: add-01

Additional information

Used simplifier rules: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i; Used terms: j0, i0 * 8