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Proof step 1 from proof uint2bits-len (conclusion)

0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1) ⊦ # uint2bits(j) = i→n(i + 2)

User applied rule induction

And obtained the following new goals

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

Rule argument

indvar: i→n(j) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j)