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Let x,y,zϵB, where B is a Boolean algebra. Simplify (x∧y)∨(x^'∧y∧z^')∨(y∧z) As much as possible.

Let x,y,zϵB, where B is a Boolean algebra. Simplify (x∧y)∨(x^'∧y∧z^')∨(y∧z) As much as possible.

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Answer #1

Given expression is : (αΛy) ντΛyΛ2)ν(yΛ2)

= (αΛy) ντΛ Λy) ν yΛ2)    [ Commutative property]

= [(a VX AZ) Ay] V (y Az)

= TV 2) A (TV :) Ay] V (YA 2)

= TIA(RV:) A y vy A2) [ I is the identity element]

= [(x\vee z')\wedge y]\vee(y\wedge z)

= (x\wedge y)\vee (z'\wedge y)\vee(y\wedge z)

= (αΛy) ν yΛ2)ν(yΛ2) [ Commutative property]

= (x\wedge y)\vee [y\wedge (z'\vee z)]

= (x\wedge y)\vee (y\wedge I)

= (x\wedge y)\vee (I\wedge y)

= (x\vee I)\wedge y

= I\wedge y

= y

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