For a valid chain of deduction in boolean algebra, does the chain of its dual hold as well?

Question

For example,

xy + x'z + yz

= xy + x'z + yz(x+x')

= xy + x'z + yzx + yzx'

= xy + xyz + x'z + x'zy

= xy(1+z) + x'z(1+y)

= xy + x'z

Hence xy + x'z + yz = xy + x'z. Also its dual must hold as well. (x+y) (x'+z) (y+z)= (x+y) (x'+z)

Could this dual result be derived directly using the dual of the whole chain given above? The dual of the whole chain is:

(x+y) (x'+z) (y+z)

= (x+y) (x'+z) (y+z) + xx'

= (x+y) (x'+z) (y+z+x) (y+z+x')

= (x+y) (x+y+z) (x'+z) (x'+z+y)

= (x+y) + (0.z) (x'+z) + (0.y)

= (x+y) (x'+z)

Point of interest is that the two bold lines does violate precedence of . over +. Still, since dual of each step holds, is the chain of deduction valid?

Answer 1

Yes. You could see this from seeing common axiom sets for a Boolean Algebra and noting the duality of the axioms. But, why it works out that way requires deeper digging, because there exist many different axiom sets for a Boolean Algebra.

The reason why it works lies in that there exists an isomorphism between the equational sub-structures (B, +, ') and (B, . , ').

Answer 2

For equations like this, yes.

Note that the "precedence" of $\cdot$ over $+$ is just a convention, and to be perfectly accurate we should write parentheses at all times... except it becomes impossible to read/understand.

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One small correction, however, the second-last line should be $$\color{red}{(}( x + y ) + ( 0 \cdot z ) \color{red}{)} \cdot \color{red}{(} ( x^\prime + z ) + ( 0 \cdot y ) \color{red}{)}$$ or perhaps just $$\color{red}{(} x + y + ( 0 \cdot z ) \color{red}{)} \cdot \color{red}{(} x^\prime + z + ( 0 \cdot y ) \color{red}{)}.$$