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Proof step 5 from proof uint2bits-2pot-03 (
previous step
)
¬ true ∈ bits, bits2uint(uint2bits(i)) = 2 ^ n→i(# bits), uint2bits(i) = false ' + true ' + bits, false ∈ uint2bits(i), 0 ≤ i ⊦ i = 2 ^ n→i(# bits), true ∈ bits
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 6
:
⊦ 0 ≤ i → bits2uint(uint2bits(i)) = i
Rule argument
name: uint2uint-pos rotate: F subst: [i] → [i]
Additional information
Used terms: i