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

¬ 0 ≤ ~(j * 2 ^ i + j0), 0 ≤ i, j < 0, j0 < 0, 0 < 2 ^ i + j0 ⊦ # int2bits(~(j * 2 ^ i + j0)) = # int2bits(~ j) + i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i, j < 0, j0 < 0, 0 < 2 ^ i + j0, 0 < j * 2 ^ i + j0 ⊦ # int2bits(~(j * 2 ^ i + j0)) = # int2bits(~ j) + i→n(i)

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;