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Proof step 6 from proof int2bits-len-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j = 2 ^ n→i(# int2bits(j)), j < 0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;