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Proof step 1 from proof int2bits-len-pos-03 (conclusion)

⊦ j < 0 ∧ 0 ≤ i → # int2bits(~(j * 2 ^ i)) = # int2bits(~ j) + i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j < 0, 0 ≤ i ⊦ # int2bits(~(j * 2 ^ i)) = # int2bits(~ j) + i→n(i)

Rule argument

Additional information

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