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Proof step 26 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: bits ≠ [] ⊦ int2bits(bits2int(bits)) = sgn(bits) subst: [bits] → [bitsand(int2bits(i), int2bits(j))] rotate: F spec: int2bits inst: name: bits2bits-def

Additional information

Used simplifier rules: ⊦ bitsand(int2bits(i), int2bits(i0)) ≠ []; Used terms: bitsand(int2bits(i), int2bits(j))