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Proof step 48 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 49:
⊦ # bits ≤ n → # fillsgn(n, bits) = n

Rule argument

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

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n → ¬ n < m; Used terms: int2bits(i), 8