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Proof step 168 from proof byte-law-and-one (previous step)

7 = # fillsgn(7, int2bits(i)), # bits1 = # int2bits(i), int2bits(j) = (bits0 + bits2) + bits1, # bits2 + # bits1 = 8, # bits1 < 8, # bits1 < # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), i < 0, 8 ≤ # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), 8 ≤ # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) ⊦ bits2 = [], bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) = [], lastn(8, fillsgn(8, (bits0 + bits2) + bitsand(bits1, int2bits(i)))) = lastn(8, bitsand(bits2 + bits1, fillsgn(8, int2bits(i))))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ n ≤ # bits → fillsgn(n, bits) = bits subst: [bits, n] → [(bits0 + bits2) + bitsand(bits1, int2bits(i)), 8] rotate: F spec: bitopsaux inst: name: fillsgn-longer

Additional information

Used terms: (bits0 + bits2) + bitsand(bits1, int2bits(i)), 8