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Proof step 2 from proof int2bits-len-mon (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ j < 0 → abs(j) = 0 - j
Proof step 4:
¬ i < 0, # int2bits(j) < # int2bits(i), abs(i) < abs(j) ⊦
Proof step 24:
i < 0, # int2bits(j) < # int2bits(i), 0 - i < abs(j) ⊦

Rule argument

seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [i] rotate: F spec: int-abs inst: name: Abs-neg

Additional information

Used terms: i