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Proof step 24 from proof int2bits-len-mon (previous step)

i < 0, # int2bits(j) < # int2bits(i), 0 - i < abs(j) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
1 < abs(j), i < 0, # int2bits(j) < # int2bits(i), 0 < abs(j) + i ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ abs(i);% ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)% ⊦ 0 < abs(i) ↔ i ≠ 0;% ⊦ 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 < j + i;% ⊦ 0 - i = ~ i;