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Proof step 38 from proof int2bits-len-mon (previous step)

j < 0, # int2bits(j) < # int2bits(~(i + 1)), i < 0, 1 < 0 - j, 0 < (0 - j) + i ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
j < 0, j < i, i < 0, # int2bits(j) < # int2bits(~(i + 1)), j < ~1 ⊦ j = 0

Rule argument

Additional information

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