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Proof step 2 from proof int2bits-less-2pot (previous step)

0 ≤ j, 2 ^ (n→i(# int2bits(j)) - 1) ≤ j ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i ∧ i ≤ j → # uint2bits(i) ≤ # uint2bits(j)
Proof step 4:
0 ≤ 2 ^ (n→i(# int2bits(j)) - 1) ∧ 2 ^ (n→i(# int2bits(j)) - 1) ≤ j → # uint2bits(2 ^ (n→i(# int2bits(j)) - 1)) ≤ # uint2bits(j), 0 ≤ j, 2 ^ (n→i(# int2bits(j)) - 1) ≤ j ⊦

Rule argument

name: uint2bits-len-mon subst: [i, j] → [2 ^ (n→i(# int2bits(j)) - 1), j]

Additional information

Used terms: 2 ^ (n→i(# int2bits(j)) - 1), j