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Simplify using boolean algebra laws/formulas
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$begingroup$
I am trying to learn logic expression simplification using boolean algebra laws/formulas, but I don't understand it at all. We have this expression:
$ (x'wedge y wedge z' ) vee (x' wedge z) vee (x wedge y) $ . In an example solution, we did this:
= $ (x'wedge y wedge z' ) vee (x' wedge z) vee (x wedge y) $
= $ (x'wedge y wedge z' ) vee (x' wedge y wedge z) vee (x' wedge y' wedge z ) vee (x wedge y) $
= $ (x'wedge y ) vee (x' wedge z ) vee (x wedge y) $
= $ y vee (x' wedge z ) $
However, I don't understand those steps at all. Could someone explain to me what boolean algebra laws/formulas we did apply at those steps and how? I would be grateful. Thanks!
logic boolean-algebra
asked Mar 17 at 16:32
J. DoeJ. Doe
1
$endgroup$
add a comment |
$begingroup$
I am trying to learn logic expression simplification using boolean algebra laws/formulas, but I don't understand it at all. We have this expression:
$ (x'wedge y wedge z' ) vee (x' wedge z) vee (x wedge y) $ . In an example solution, we did this:
= $ (x'wedge y wedge z' ) vee (x' wedge z) vee (x wedge y) $
= $ (x'wedge y wedge z' ) vee (x' wedge y wedge z) vee (x' wedge y' wedge z ) vee (x wedge y) $
= $ (x'wedge y ) vee (x' wedge z ) vee (x wedge y) $
= $ y vee (x' wedge z ) $
However, I don't understand those steps at all. Could someone explain to me what boolean algebra laws/formulas we did apply at those steps and how? I would be grateful. Thanks!
logic boolean-algebra
asked Mar 17 at 16:32
J. DoeJ. Doe
1
$endgroup$
$begingroup$
The identity $a equiv (awedge b) vee (a wedge b')$ is exploited in all lines. On 2nd row, it is used on $(x'wedge z)$, and the term $x'wedge y wedge z$ is also duplicated to use the identity twice on 3rd row.
$endgroup$
– FormerMath
Mar 17 at 17:41
add a comment |
$begingroup$
I am trying to learn logic expression simplification using boolean algebra laws/formulas, but I don't understand it at all. We have this expression:
$ (x'wedge y wedge z' ) vee (x' wedge z) vee (x wedge y) $ . In an example solution, we did this:
= $ (x'wedge y wedge z' ) vee (x' wedge z) vee (x wedge y) $
= $ (x'wedge y wedge z' ) vee (x' wedge y wedge z) vee (x' wedge y' wedge z ) vee (x wedge y) $
= $ (x'wedge y ) vee (x' wedge z ) vee (x wedge y) $
= $ y vee (x' wedge z ) $
However, I don't understand those steps at all. Could someone explain to me what boolean algebra laws/formulas we did apply at those steps and how? I would be grateful. Thanks!
logic boolean-algebra
asked Mar 17 at 16:32
J. DoeJ. Doe
1
$endgroup$
I am trying to learn logic expression simplification using boolean algebra laws/formulas, but I don't understand it at all. We have this expression:
$ (x'wedge y wedge z' ) vee (x' wedge z) vee (x wedge y) $ . In an example solution, we did this:
= $ (x'wedge y wedge z' ) vee (x' wedge z) vee (x wedge y) $
= $ (x'wedge y wedge z' ) vee (x' wedge y wedge z) vee (x' wedge y' wedge z ) vee (x wedge y) $
= $ (x'wedge y ) vee (x' wedge z ) vee (x wedge y) $
= $ y vee (x' wedge z ) $
However, I don't understand those steps at all. Could someone explain to me what boolean algebra laws/formulas we did apply at those steps and how? I would be grateful. Thanks!
logic boolean-algebra
logic boolean-algebra
asked Mar 17 at 16:32
J. DoeJ. Doe
1
asked Mar 17 at 16:32
J. DoeJ. Doe
1
asked Mar 17 at 16:32
J. DoeJ. Doe
1
asked Mar 17 at 16:32
J. DoeJ. Doe
1
asked Mar 17 at 16:32
J. DoeJ. Doe
1
1
$begingroup$
The identity $a equiv (awedge b) vee (a wedge b')$ is exploited in all lines. On 2nd row, it is used on $(x'wedge z)$, and the term $x'wedge y wedge z$ is also duplicated to use the identity twice on 3rd row.
$endgroup$
– FormerMath
Mar 17 at 17:41
add a comment |
$begingroup$
The identity $a equiv (awedge b) vee (a wedge b')$ is exploited in all lines. On 2nd row, it is used on $(x'wedge z)$, and the term $x'wedge y wedge z$ is also duplicated to use the identity twice on 3rd row.
$endgroup$
– FormerMath
Mar 17 at 17:41
$begingroup$
The identity $a equiv (awedge b) vee (a wedge b')$ is exploited in all lines. On 2nd row, it is used on $(x'wedge z)$, and the term $x'wedge y wedge z$ is also duplicated to use the identity twice on 3rd row.
$endgroup$
– FormerMath
Mar 17 at 17:41
$begingroup$
The identity $a equiv (awedge b) vee (a wedge b')$ is exploited in all lines. On 2nd row, it is used on $(x'wedge z)$, and the term $x'wedge y wedge z$ is also duplicated to use the identity twice on 3rd row.
$endgroup$
– FormerMath
Mar 17 at 17:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The following general equivalence principles are used:
Adjacency
$p = (p land q) lor (p land q')$
$p = (p lor q) land (p lor q')$
Idempotence
$p = p lor p$
$p = p land p$
Applied to your statement:
$ (x'land y land z' ) lor (x' land z) lor (x land y) overset{Adjacency: x' land z = (x' land y land z) lor (x' land y' land z)}{=}$
$ (x'land y land z' ) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Idempotence: (x' land y land z) = (x' land y land z) lor (x' land y land z)}{=}$
$ (x'land y land z' ) lor (x' land y land z) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Adjacency: (x'land y land z' ) lor (x' land y land z) = x' land y}{=}$
$ (x'land y) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Adjacency: (x' land y land z) lor (x' land y' land z) = x' land z}{=}$
$ (x'land y) lor (x' land z) lor (x land y) overset{Adjacency: (x'land y) lor (x land y) = y}{=}$
$y lor (x' land z)$
answered Mar 17 at 19:30
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The following general equivalence principles are used:
Adjacency
$p = (p land q) lor (p land q')$
$p = (p lor q) land (p lor q')$
Idempotence
$p = p lor p$
$p = p land p$
Applied to your statement:
$ (x'land y land z' ) lor (x' land z) lor (x land y) overset{Adjacency: x' land z = (x' land y land z) lor (x' land y' land z)}{=}$
$ (x'land y land z' ) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Idempotence: (x' land y land z) = (x' land y land z) lor (x' land y land z)}{=}$
$ (x'land y land z' ) lor (x' land y land z) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Adjacency: (x'land y land z' ) lor (x' land y land z) = x' land y}{=}$
$ (x'land y) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Adjacency: (x' land y land z) lor (x' land y' land z) = x' land z}{=}$
$ (x'land y) lor (x' land z) lor (x land y) overset{Adjacency: (x'land y) lor (x land y) = y}{=}$
$y lor (x' land z)$
answered Mar 17 at 19:30
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
$begingroup$
The following general equivalence principles are used:
Adjacency
$p = (p land q) lor (p land q')$
$p = (p lor q) land (p lor q')$
Idempotence
$p = p lor p$
$p = p land p$
Applied to your statement:
$ (x'land y land z' ) lor (x' land z) lor (x land y) overset{Adjacency: x' land z = (x' land y land z) lor (x' land y' land z)}{=}$
$ (x'land y land z' ) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Idempotence: (x' land y land z) = (x' land y land z) lor (x' land y land z)}{=}$
$ (x'land y land z' ) lor (x' land y land z) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Adjacency: (x'land y land z' ) lor (x' land y land z) = x' land y}{=}$
$ (x'land y) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Adjacency: (x' land y land z) lor (x' land y' land z) = x' land z}{=}$
$ (x'land y) lor (x' land z) lor (x land y) overset{Adjacency: (x'land y) lor (x land y) = y}{=}$
$y lor (x' land z)$
answered Mar 17 at 19:30
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
$begingroup$
The following general equivalence principles are used:
Adjacency
$p = (p land q) lor (p land q')$
$p = (p lor q) land (p lor q')$
Idempotence
$p = p lor p$
$p = p land p$
Applied to your statement:
$ (x'land y land z' ) lor (x' land z) lor (x land y) overset{Adjacency: x' land z = (x' land y land z) lor (x' land y' land z)}{=}$
$ (x'land y land z' ) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Idempotence: (x' land y land z) = (x' land y land z) lor (x' land y land z)}{=}$
$ (x'land y land z' ) lor (x' land y land z) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Adjacency: (x'land y land z' ) lor (x' land y land z) = x' land y}{=}$
$ (x'land y) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Adjacency: (x' land y land z) lor (x' land y' land z) = x' land z}{=}$
$ (x'land y) lor (x' land z) lor (x land y) overset{Adjacency: (x'land y) lor (x land y) = y}{=}$
$y lor (x' land z)$
answered Mar 17 at 19:30
Bram28Bram28
63.9k44793
$endgroup$
The following general equivalence principles are used:
Adjacency
$p = (p land q) lor (p land q')$
$p = (p lor q) land (p lor q')$
Idempotence
$p = p lor p$
$p = p land p$
Applied to your statement:
$ (x'land y land z' ) lor (x' land z) lor (x land y) overset{Adjacency: x' land z = (x' land y land z) lor (x' land y' land z)}{=}$
$ (x'land y land z' ) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Idempotence: (x' land y land z) = (x' land y land z) lor (x' land y land z)}{=}$
$ (x'land y land z' ) lor (x' land y land z) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Adjacency: (x'land y land z' ) lor (x' land y land z) = x' land y}{=}$
$ (x'land y) lor (x' land y land z) lor (x' land y' land z) lor (x land y) overset{Adjacency: (x' land y land z) lor (x' land y' land z) = x' land z}{=}$
$ (x'land y) lor (x' land z) lor (x land y) overset{Adjacency: (x'land y) lor (x land y) = y}{=}$
$y lor (x' land z)$
answered Mar 17 at 19:30
Bram28Bram28
63.9k44793
answered Mar 17 at 19:30
Bram28Bram28
63.9k44793
answered Mar 17 at 19:30
Bram28Bram28
63.9k44793
answered Mar 17 at 19:30
Bram28Bram28
63.9k44793
63.9k44793
add a comment |
add a comment |
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$begingroup$
The identity $a equiv (awedge b) vee (a wedge b')$ is exploited in all lines. On 2nd row, it is used on $(x'wedge z)$, and the term $x'wedge y wedge z$ is also duplicated to use the identity twice on 3rd row.
$endgroup$
– FormerMath
Mar 17 at 17:41