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Proof step 21 from proof int2bits-len-mon (previous step)

0 ≤ ~(j + 1), 0 ≤ i, j < 0, j < i, i + j < 0, # uint2bits(~(j + 1)) < # uint2bits(i) ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
0 ≤ i, j < i, i + j < 0, # uint2bits(~(j + 1)) < # uint2bits(i), j < 0 ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;