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Proof step 30 from proof int2bits-len-mon (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
i < j, 0 < j + i, i < 0, # int2bits(j) < # int2bits(~(i + 1)), 0 ≤ j, 1 < j, i + 1 < j, 0 < j ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i < i + j ↔ 0 < j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i +1 = i + 1;% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)% ⊦ i < j → ¬ j < i;% ⊦ ¬ j < 0 → abs(j) = j;% ⊦ ¬ i < j → (i ≠ j ↔ j < i);