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Proof step 10 from proof int2bits-len-mon (previous step)

¬ j < 0, # int2bits(j) < # uint2bits(i), i < abs(j), 0 ≤ i ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ i, # int2bits(j) < # uint2bits(i), i < j, 0 < j ⊦

Rule argument

Additional information

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