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Proof step 4 from proof uint2bits-2pot-04 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 5:
bits2uint(uint2bits(i)) = bits2uint(false ' + true ' + bits), uint2bits(i) = false ' + true ' + bits, false ∈ uint2bits(i), 0 ≤ i ⊦ 2 ^ (n→i(# bits) + 1) = i + 1, false ∈ bits
Proof step 8:
uint2bits(i) = false ' + true ' + bits, false ∈ uint2bits(i), 0 ≤ i ⊦ bits2uint(uint2bits(i)) = bits2uint(false ' + true ' + bits), 2 ^ (n→i(# bits) + 1) = i + 1, false ∈ bits

Rule argument

fma: bits2uint(uint2bits(i)) = bits2uint(false ' + true ' + bits)

Additional information

Used formulas: bits2uint(uint2bits(i)) = bits2uint(false ' + true ' + bits)