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Proof step 5 from proof int2bits-restsgn (
previous step
)
0 ≤ j, 2 ^ n→i(# int2bits(j)) < j ⊦ sgn(int2bits(2 ^ n→i(# int2bits(j)) - j).rest) = int2bits(2 ^ n→i(# int2bits(j)) - j).rest, ∃ i. 2 ^ i = j
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 6
:
⊦ 0 ≤ j → ¬ 2 ^ n→i(# int2bits(j)) < j
Rule argument
name: int2bits-less-2pot-02 rotate: F subst: [j] → [j]
Additional information
Used terms: j