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Proof step 1 from proof int2bits-len-mon (conclusion)

⊦ abs(i) < abs(j) → # int2bits(i) ≤ # int2bits(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# int2bits(j) < # int2bits(i), abs(i) < abs(j) ⊦

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;