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Proof step 9 from proof int2bits-lslen-07 (previous step)

i→n(i + 2) + 1 < # int2bits(j + 1), j < 0, j + 2 ^ i < 0, 0 ≤ i, 0 ≤ 2 ^ i * 2 + j ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
0 ≤ i, 0 ≤ j ⊦ i→n(i + j) = i→n(i) + i→n(j)
Proof step 11:
0 ≤ i ∧ 0 ≤ 2, (i→n(i) + i→n(2)) + 1 < # int2bits(j + 1), j < 0, j + 2 ^ i < 0, 0 ≤ i, 0 ≤ 2 ^ i * 2 + j ⊦

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 + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false); Used terms: 2, i