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Proof step 20 from proof byte-law-and-one (previous step)

bitsand(sgn(lastn(8, fillsgn(8, int2bits(i)))), int2bits(j)) ≠ [], lastn(8, fillsgn(8, int2bits(i))) ≠ [] ⊦ band(i, j) = bits2int(bitsand(sgn(lastn(8, fillsgn(8, int2bits(i)))), int2bits(j))), b2i(i2b(band(i, j))) = bits2int(lastn(8, fillsgn(8, sgn(bitsand(sgn(lastn(8, fillsgn(8, int2bits(i)))), int2bits(j))))))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

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