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Proof step 14 from proof int2bits-len-add (previous step)

0 < ~ j ∧ 0 ≤ ~ j0 ∧ ~ j0 < 2 ^ i ∧ 0 ≤ i, 0 ≤ i, j < 0, j0 < 0, 0 < 2 ^ i + j0, j * 2 ^ i + j0 ≤ 0 ⊦ # uint2bits(~ j) + i→n(i) = # int2bits(~ j) + i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
0 ≤ i, j < 0, j0 < 0, j * 2 ^ i + j0 ≤ 0, 0 < 2 ^ i + j0 ⊦ # uint2bits(~ j) = # int2bits(~ j)

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i < j ↔ 0 < j + i;% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ 0 < ~ i ↔ i < 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;