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Proof step 195 from proof byte-law-and-one (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 196:
n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n
Proof step 197:
int2bits(i) = bits0 + bits, # bits = 8 ⊦ 8 ≤ # int2bits(j), bitsand(bits, int2bits(j)) = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, int2bits(j)))) = lastn(8, fillsgn(8, bitsand(bits, int2bits(j))))
Proof step 224:
∃ bits0, bits1. int2bits(j) = bits0 + bits1 ∧ # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8 ⊦ bitsand(bits, int2bits(j)) = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, int2bits(j)))) = lastn(8, fillsgn(8, bitsand(bits, int2bits(j))))

Rule argument

seq: n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n subst: [n, bits] → [8, int2bits(j)] rotate: F spec: list inst: bitlista name: split-rest

Additional information

Used terms: 8, int2bits(j)