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Proof step 78 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ # bits ≤ n → # fillsgn(n, bits) = n subst: [bits, n] → [bitsand(fillsgn(8, int2bits(i)), int2bits(j)), 8] rotate: F spec: bitopsaux inst: name: fill-length

Additional information

Used terms: bitsand(fillsgn(8, int2bits(i)), int2bits(j)), 8