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Proof step 8 from proof int2bits-less-2pot-01 (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 9:
⊦ abs(i) < abs(j) → # int2bits(i) ≤ # int2bits(j)
Proof step 10:
abs(2 ^ (n→i(# int2bits(j)) - 1)) < abs(j) → # int2bits(2 ^ (n→i(# int2bits(j)) - 1)) ≤ # int2bits(j), j < 0, 2 ^ (n→i(# int2bits(j)) - 1) + j < 0 ⊦

Rule argument

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

Additional information

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