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Proof step 2 from proof uint2bits-2pot-04 (previous step)

0 ≤ i ⊦ uint2bits(i) = false ' + true ' + bits ↔ 2 ^ (n→i(# bits) + 1) = i + 1, false ∈ bits

User applied rule case distinction

And obtained the following new goals

Proof step 3:
uint2bits(i) = false ' + true ' + bits, 0 ≤ i ⊦ 2 ^ (n→i(# bits) + 1) = i + 1, false ∈ bits
Proof step 9:
2 ^ (n→i(# bits) + 1) = i + 1, 0 ≤ i ⊦ uint2bits(i) = false ' + true ' + bits, false ∈ bits

Rule argument

pos: right 1

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