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Proof step 27 from proof int2bits-len-mon (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
# int2bits(j) < # int2bits(~(i + 1)), i < 0, 1 < abs(j), 0 < abs(j) + i ⊦ j = 0

Rule argument

Additional information

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