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Proof step 4 from proof int2bits-len-eq (previous step)

0 < j → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)), j = 2 ^ n→i(# int2bits(j)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
j = 2 ^ n→i(# int2bits(j)), ¬ j ≤ 0 → 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;