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Proof step 110 from proof byte-law-and-one (previous step)

# bits = 8, int2bits(j) = bits0 + bits, 7 = # fillsgn(7, int2bits(i)), # int2bits(i) < 8, 0 ≤ i ⊦ 8 = # fillsgn(8, bitsand(bits, int2bits(i))), bits0 + bits = [], lastn(8, fillsgn(8, bitsand(bits, int2bits(i)))) = lastn(8, fillsgn(8, bitsand(bits, false ' + fillsgn(7, int2bits(i)))))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 111:
⊦ # bits ≤ n → # fillsgn(n, bits) = n
Proof step 112:
¬ # bitsand(bits, int2bits(i)) ≤ 8, # bits = 8, int2bits(j) = bits0 + bits, 7 = # fillsgn(7, int2bits(i)), # int2bits(i) < 8, 0 ≤ i ⊦ 8 = # fillsgn(8, bitsand(bits, int2bits(i))), bits0 + bits = [], lastn(8, fillsgn(8, bitsand(bits, int2bits(i)))) = lastn(8, fillsgn(8, bitsand(bits, false ' + fillsgn(7, int2bits(i)))))

Rule argument

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

Additional information

Used terms: bitsand(bits, int2bits(i)), 8