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Proof step 5 from proof int2int-lastn-negneg (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 6:
bits = [], # bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), m ≤ # int2bits(i), lastn(m, int2bits(i)).first, i < 0 ⊦ i + 2 ^ n→i(m) = 0, bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m), m = 0
Proof step 12:
# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), m ≤ # int2bits(i), lastn(m, int2bits(i)).first, i < 0 ⊦ bits = [], i + 2 ^ n→i(m) = 0, bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m), m = 0

Rule argument

fma: bits = []

Additional information

Used formulas: bits = []