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Proof step 39 from proof byte-law-and-one (previous step)

⊦ bitsand(lastn(8, fillsgn(8, int2bits(i))), int2bits(j)) = [], lastn(8, fillsgn(8, sgn(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 40:
⊦ bits ≠ [] → lastn(n, fillsgn(n, sgn(bits))) = lastn(n, fillsgn(n, bits))
Proof step 41:
bitsand(int2bits(i), int2bits(j)) ≠ [] ⊦ 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))))

Rule argument

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

Additional information

Used simplifier rules: ⊦ bitsand(int2bits(i), int2bits(i0)) ≠ []; Used terms: bitsand(int2bits(i), int2bits(j)), 8