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Proof step 225 from proof byte-law-and-one (previous step)

int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8 ⊦ bitsand(bits, bits2 + bits1) = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, bits2 + bits1))) = lastn(8, fillsgn(8, bitsand(bits, bits2 + bits1)))

User applied rule cut formula

And obtained the following new goals

Proof step 226:
int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8 ⊦ bits0 = [], bitsand(bits, bits2 + bits1) = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, bits2 + bits1))) = lastn(8, fillsgn(8, bitsand(bits, bits2 + bits1)))

Rule argument

fma: bits0 = []

Additional information

Used simplifier rules: ⊦ [] + bits = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used formulas: bits0 = []