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Proof step 28 from proof int2bits-lslen-08 (previous step)

2 ^ i * 2 < 2 ^ (2 + i), 0 ≤ i, 3 + i→n(i) < i→n(3 + i) ⊦ ~3 = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
0 ≤ i, 0 ≤ j ⊦ i→n(i + j) = i→n(i) + i→n(j)

Rule argument

seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i + j) = i→n(i) + i→n(j) subst: [j, i] → [i, 3] rotate: F spec: intnat inst: name: add-01

Additional information

Used simplifier rules: ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);% ⊦ 0 ≤ i → abs(i) = i;% ⊦ ¬ i < i;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: i, 3