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Proof step 278 from proof byte-law-and-one (previous step)

int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8, bits.first ⊦ lastn(8, bitsand(bits1, bits)) = lastn(8, fillsgn(8, bits2 + bitsand(bits1, bits))), bits2 + bits1 = [], bits2 = [], bits0 = [], bits1 = [], bits = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 279:
⊦ n ≤ # bits → fillsgn(n, bits) = bits
Proof step 280:
¬ 8 ≤ #(bits2 + bitsand(bits1, bits)), int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8, bits.first ⊦ lastn(8, bitsand(bits1, bits)) = lastn(8, fillsgn(8, bits2 + bitsand(bits1, bits))), bits2 + bits1 = [], bits2 = [], bits0 = [], bits1 = [], bits = []
Proof step 283:
8 ≤ #(bits2 + bitsand(bits1, bits)), int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8, bits.first ⊦ lastn(8, bitsand(bits1, bits)) = lastn(8, bits2 + bitsand(bits1, bits)), bits2 + bits1 = [], bits2 = [], bits0 = [], bits1 = [], bits = []

Rule argument

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

Additional information

Used terms: bits2 + bitsand(bits1, bits), 8