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Proof step 9 from proof uint2bits-sgn-2pot (previous step)

uint2bits(j) = false ' + true ' + bits, 0 < j, ∀ i. ¬ i < 0 → j ≠ 2 ^ i ⊦ sgn(true ' + bits) = true ' + true ' + bits, true ∈ bits

User applied rule cut formula

And obtained the following new goals

Proof step 10:
bits2uint(uint2bits(j)) = bits2uint(false ' + true ' + bits), uint2bits(j) = false ' + true ' + bits, 0 < j, ∀ i. ¬ i < 0 → j ≠ 2 ^ i ⊦ sgn(true ' + bits) = true ' + true ' + bits, true ∈ bits

Rule argument

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

Additional information

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