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Proof step 6 from proof int2int-lastn-negpos (previous step)

# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), i < 0, m ≤ # int2bits(i) ⊦ bits = [], bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m) + 2 ^ n→i(m), i % 2 ^ n→i(m) = 0, m = 0, lastn(m, int2bits(i)).first

User applied rule cut formula

And obtained the following new goals

Proof step 7:
bits2int(int2bits(i)) = bits2int(bits + lastn(m, int2bits(i))), # bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), i < 0, m ≤ # int2bits(i) ⊦ bits = [], bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m) + 2 ^ n→i(m), i % 2 ^ n→i(m) = 0, m = 0, lastn(m, int2bits(i)).first

Rule argument

fma: bits2int(int2bits(i)) = bits2int(bits + lastn(m, int2bits(i)))

Additional information

Used simplifier rules: int2int-inj: ⊦ bits2int(int2bits(i)) = i; Used formulas: bits2int(int2bits(i)) = bits2int(bits + lastn(m, int2bits(i)))