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Proof step 89 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: ⊦ fillsgn(n, bits) = [] ↔ bits = [] ∧ n = 0;% ⊦ int2bits(j) ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits; Used terms: bits0 + bits, fillsgn(8, int2bits(i))