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Proof step 134 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 135:
⊦ bits ≠ [] ∧ bits0 ≠ [] → fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0))
Proof step 136:
int2bits(i) ≠ [] ∧ bits ≠ [], # bits = 8, int2bits(j) = bits0 + bits, 7 = # fillsgn(7, int2bits(i)), 8 = # fillsgn(8, bitsand(bits, int2bits(i))), 8 = # fillsgn(8, bitsand(int2bits(i), bits)), # int2bits(i) < 8, 0 ≤ i ⊦ bits = [], fillsgn(8, int2bits(i)) = [], bits0 + bits = [], fillsgn(8, bitsand(bits, int2bits(i))) = fillsgn(8, bitsand(int2bits(i), bits))

Rule argument

seq: ⊦ bits ≠ [] ∧ bits0 ≠ [] → fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0)) subst: [bits0, bits, n] → [bits, int2bits(i), 8] rotate: F spec: intops inst: name: bitsand-fillsgn

Additional information

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