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Proof step 5 from proof uint2bits-len-min (previous step)

0 < j, 0 ≤ i, j < 2 ^ (i0 + 1), 2 ^ i0 ≤ j, # uint2bits(j) ≤ i→n(i) + 1, 2 ^ i ≤ j, 1 < 2 ^ (i0 + 1), i ≤ i0, 0 ≤ i0, 0 < 2 ^ (i0 + 1), 0 ≤ 2 ^ (i0 + 1) ⊦ 0 = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1) ⊦ # uint2bits(j) = i→n(i + 2)
Proof step 7:
0 ≤ i0 ∧ 2 ^ i0 ≤ j ∧ j < 2 ^ (i0 + 1), 0 < j, 0 ≤ i, j < 2 ^ (i0 + 1), 2 ^ i0 ≤ j, i→n(i0 + 2) ≤ i→n(i) + 1, 2 ^ i ≤ j, 1 < 2 ^ (i0 + 1), i ≤ i0, 0 ≤ i0, 0 < 2 ^ (i0 + 1), 0 ≤ 2 ^ (i0 + 1) ⊦ 0 = j

Rule argument

name: uint2bits-len rotate: F subst: [i, j] → [i0, j]

Additional information

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