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Proof step 25 from proof int2bits-lslen-08 (previous step)

i→n(i + 2) + 1 < i→n(3 + i), 2 ^ i * 2 < 2 ^ (2 + i), 0 ≤ i ⊦ ~3 = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
0 ≤ i, 0 ≤ j ⊦ i→n(i + j) = i→n(i) + i→n(j)
Proof step 27:
0 ≤ i ∧ 0 ≤ 2, (i→n(i) + i→n(2)) + 1 < i→n(3 + i), 2 ^ i * 2 < 2 ^ (2 + i), 0 ≤ i ⊦ ~3 = i

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); Used terms: 2, i