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Proof step 59 from proof byte-law-and-one (previous step)

8 = # fillsgn(8, int2bits(i)), # int2bits(i) < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < 8, # int2bits(i) < 8, 8 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) + 1 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) + 1 < # bitsand(int2bits(i), int2bits(j)) ⊦ lastn(8, bitsand(int2bits(i), int2bits(j))) = lastn(8, fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j)))), 8 = # fillsgn(8, bitsand(int2bits(i), int2bits(j))), bitsand(fillsgn(8, int2bits(i)), int2bits(j)) = []

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: # int2bits(i) < # int2bits(j)

Additional information

Used formulas: # int2bits(i) < # int2bits(j)