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Proof step 24 from proof uint2bits-len-01 (previous step)

# uint2bits(j / 2) = i→n(2 + i), 0 ≤ i, j < 3 * 2 ^ i, 2 * 2 ^ i ≤ j, 0 < j, 0 < 2 ^ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ j0 < 3 * 2 ^ i ∧ 2 * 2 ^ i ≤ j0 ∧ 0 < 2 ^ i → # uint2bits(j0) = i→n(i + 3) ⊦ i→n(2 + i) + 1 = i→n(i + 3), j = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
0 ≤ i, 0 ≤ j ⊦ i→n(i + j) = i→n(i) + i→n(j)
Proof step 26:
0 ≤ i ∧ 0 ≤ 2, # uint2bits(j / 2) = i→n(i) + i→n(2), 0 ≤ i, j < 3 * 2 ^ i, 2 * 2 ^ i ≤ j, 0 < j, 0 < 2 ^ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ j0 < 3 * 2 ^ i ∧ 2 * 2 ^ i ≤ j0 ∧ 0 < 2 ^ i → # uint2bits(j0) = i→n(i + 3) ⊦ (i→n(i) + i→n(2)) + 1 = i→n(i + 3), j = 1

Rule argument

seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i + j) = i→n(i) + i→n(j) subst: [j, i] → [2, i] rotate: F spec: intnat inst: name: add-01

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i + j = j + i; Used terms: 2, i