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Proof step 86 from proof byte-law-and-one (previous step)

8 = # fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j))), 8 = # fillsgn(8, bitsand(int2bits(i), int2bits(j))), 8 = # fillsgn(8, int2bits(i)), # int2bits(j) < 8, # int2bits(i) < 8 ⊦ bitsand(fillsgn(8, int2bits(i)), int2bits(j)) = [], fillsgn(8, bitsand(int2bits(i), int2bits(j))) = fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 87:
⊦ bits ≠ [] ∧ bits0 ≠ [] → fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0))

Rule argument

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

Additional information

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