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Proof step 137 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 138:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) = bitsand(bits0, bits)

Rule argument

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

Additional information

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