[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 251 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 252:
⊦ n ≤ # bits → fillsgn(n, bits) = bits
Proof step 253:
¬ 8 ≤ #(bitsand(bits0, bits2) + bitsand(bits, bits1)), # bits = 8, int2bits(i) = bits0 + bits, # bits1 = 8, int2bits(j) = bits2 + bits1 ⊦ bits0 = [], bits2 = [], bits = [], bits2 + bits1 = [], lastn(8, fillsgn(8, bitsand(bits0, bits2) + bitsand(bits, bits1))) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits)))
Proof step 259:
8 ≤ #(bitsand(bits0, bits2) + bitsand(bits, bits1)), # bits = 8, int2bits(i) = bits0 + bits, # bits1 = 8, int2bits(j) = bits2 + bits1 ⊦ bits0 = [], bits2 = [], bits = [], bits2 + bits1 = [], lastn(8, bitsand(bits0, bits2) + bitsand(bits, bits1)) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits)))

Rule argument

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

Additional information

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