Simplifying boolean algebra expression that contains XOR
How can i simplify followed boolean algebra expression; Normally i express as simplify without XOR also this expresion contains both XOR and multiple variables.
(((A + B)' * C') ⊕ ((A' + B') * C')' ) * A
⊕ : XOR
boolean-algebra
asked Nov 11 '15 at 7:40
user289053user289053
11
add a comment |
How can i simplify followed boolean algebra expression; Normally i express as simplify without XOR also this expresion contains both XOR and multiple variables.
(((A + B)' * C') ⊕ ((A' + B') * C')' ) * A
⊕ : XOR
boolean-algebra
asked Nov 11 '15 at 7:40
user289053user289053
11
What've you tried? What rules do you know and/or are you allowed to use?
– Marnix Klooster
Nov 11 '15 at 8:39
add a comment |
How can i simplify followed boolean algebra expression; Normally i express as simplify without XOR also this expresion contains both XOR and multiple variables.
(((A + B)' * C') ⊕ ((A' + B') * C')' ) * A
⊕ : XOR
boolean-algebra
asked Nov 11 '15 at 7:40
user289053user289053
11
How can i simplify followed boolean algebra expression; Normally i express as simplify without XOR also this expresion contains both XOR and multiple variables.
(((A + B)' * C') ⊕ ((A' + B') * C')' ) * A
⊕ : XOR
boolean-algebra
boolean-algebra
asked Nov 11 '15 at 7:40
user289053user289053
11
asked Nov 11 '15 at 7:40
user289053user289053
11
asked Nov 11 '15 at 7:40
user289053user289053
11
asked Nov 11 '15 at 7:40
user289053user289053
11
asked Nov 11 '15 at 7:40
user289053user289053
11
11
What've you tried? What rules do you know and/or are you allowed to use?
– Marnix Klooster
Nov 11 '15 at 8:39
add a comment |
What've you tried? What rules do you know and/or are you allowed to use?
– Marnix Klooster
Nov 11 '15 at 8:39
What've you tried? What rules do you know and/or are you allowed to use?
– Marnix Klooster
Nov 11 '15 at 8:39
What've you tried? What rules do you know and/or are you allowed to use?
– Marnix Klooster
Nov 11 '15 at 8:39
add a comment |
2 Answers
2
active
oldest
votes
Lets say X = (A + B)' * C') and Y = ((A' + B') * C')'
We know that XOR can be represented as
$X ⊕ Y = Xoverline Y + overline X Y$
So, first simplify the inners and once you have the simplified version, you can deal with the outer A.
answered Nov 11 '15 at 11:20
cryptocrypto
1276
add a comment |
I've never been a fan of the boolean algebra notation, so this is not a direct answer to your question, but more how I would do this simplification using propositional logic notation.$
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{ref}[1]{text{(#1)}}
newcommand{then}{Rightarrow}
newcommand{followsfrom}{Leftarrow}
newcommand{true}{text{true}}
newcommand{false}{text{false}}
$
Translated into that notation, you're asked to simplify $$
big(lnot (A lor B) land lnot C ;notequiv; lnot ((lnot A lor lnot B) land C)big) ;land; A
$$
We can do that as in the following calculation:
$$calc
big(lnot (A lor B) land lnot C ;notequiv; lnot ((lnot A lor lnot B) land C)big) ;land; A
op=hint{use $;A;$ on other side of the rightmost $;land;$}
big(lnot (true lor B) land lnot C ;notequiv; lnot ((false lor lnot B) land C)big) ;land; A
op=hint{simplify}
big(false ;notequiv; lnot (lnot B land C)big) ;land; A
op=hint{simplify $;false notequiv P;$ to $;P;$; DeMorgan}
(B lor lnot C) ;land; A
endcalc$$
And now it is trivial to translate that back to boolean algebra notation...
answered Nov 12 '15 at 7:21
Marnix KloosterMarnix Klooster
4,20322146
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Lets say X = (A + B)' * C') and Y = ((A' + B') * C')'
We know that XOR can be represented as
$X ⊕ Y = Xoverline Y + overline X Y$
So, first simplify the inners and once you have the simplified version, you can deal with the outer A.
answered Nov 11 '15 at 11:20
cryptocrypto
1276
add a comment |
Lets say X = (A + B)' * C') and Y = ((A' + B') * C')'
We know that XOR can be represented as
$X ⊕ Y = Xoverline Y + overline X Y$
So, first simplify the inners and once you have the simplified version, you can deal with the outer A.
answered Nov 11 '15 at 11:20
cryptocrypto
1276
add a comment |
Lets say X = (A + B)' * C') and Y = ((A' + B') * C')'
We know that XOR can be represented as
$X ⊕ Y = Xoverline Y + overline X Y$
So, first simplify the inners and once you have the simplified version, you can deal with the outer A.
answered Nov 11 '15 at 11:20
cryptocrypto
1276
Lets say X = (A + B)' * C') and Y = ((A' + B') * C')'
We know that XOR can be represented as
$X ⊕ Y = Xoverline Y + overline X Y$
So, first simplify the inners and once you have the simplified version, you can deal with the outer A.
answered Nov 11 '15 at 11:20
cryptocrypto
1276
answered Nov 11 '15 at 11:20
cryptocrypto
1276
answered Nov 11 '15 at 11:20
cryptocrypto
1276
answered Nov 11 '15 at 11:20
cryptocrypto
1276
1276
add a comment |
add a comment |
I've never been a fan of the boolean algebra notation, so this is not a direct answer to your question, but more how I would do this simplification using propositional logic notation.$
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{ref}[1]{text{(#1)}}
newcommand{then}{Rightarrow}
newcommand{followsfrom}{Leftarrow}
newcommand{true}{text{true}}
newcommand{false}{text{false}}
$
Translated into that notation, you're asked to simplify $$
big(lnot (A lor B) land lnot C ;notequiv; lnot ((lnot A lor lnot B) land C)big) ;land; A
$$
We can do that as in the following calculation:
$$calc
big(lnot (A lor B) land lnot C ;notequiv; lnot ((lnot A lor lnot B) land C)big) ;land; A
op=hint{use $;A;$ on other side of the rightmost $;land;$}
big(lnot (true lor B) land lnot C ;notequiv; lnot ((false lor lnot B) land C)big) ;land; A
op=hint{simplify}
big(false ;notequiv; lnot (lnot B land C)big) ;land; A
op=hint{simplify $;false notequiv P;$ to $;P;$; DeMorgan}
(B lor lnot C) ;land; A
endcalc$$
And now it is trivial to translate that back to boolean algebra notation...
answered Nov 12 '15 at 7:21
Marnix KloosterMarnix Klooster
4,20322146
add a comment |
I've never been a fan of the boolean algebra notation, so this is not a direct answer to your question, but more how I would do this simplification using propositional logic notation.$
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{ref}[1]{text{(#1)}}
newcommand{then}{Rightarrow}
newcommand{followsfrom}{Leftarrow}
newcommand{true}{text{true}}
newcommand{false}{text{false}}
$
Translated into that notation, you're asked to simplify $$
big(lnot (A lor B) land lnot C ;notequiv; lnot ((lnot A lor lnot B) land C)big) ;land; A
$$
We can do that as in the following calculation:
$$calc
big(lnot (A lor B) land lnot C ;notequiv; lnot ((lnot A lor lnot B) land C)big) ;land; A
op=hint{use $;A;$ on other side of the rightmost $;land;$}
big(lnot (true lor B) land lnot C ;notequiv; lnot ((false lor lnot B) land C)big) ;land; A
op=hint{simplify}
big(false ;notequiv; lnot (lnot B land C)big) ;land; A
op=hint{simplify $;false notequiv P;$ to $;P;$; DeMorgan}
(B lor lnot C) ;land; A
endcalc$$
And now it is trivial to translate that back to boolean algebra notation...
answered Nov 12 '15 at 7:21
Marnix KloosterMarnix Klooster
4,20322146
add a comment |
I've never been a fan of the boolean algebra notation, so this is not a direct answer to your question, but more how I would do this simplification using propositional logic notation.$
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{ref}[1]{text{(#1)}}
newcommand{then}{Rightarrow}
newcommand{followsfrom}{Leftarrow}
newcommand{true}{text{true}}
newcommand{false}{text{false}}
$
Translated into that notation, you're asked to simplify $$
big(lnot (A lor B) land lnot C ;notequiv; lnot ((lnot A lor lnot B) land C)big) ;land; A
$$
We can do that as in the following calculation:
$$calc
big(lnot (A lor B) land lnot C ;notequiv; lnot ((lnot A lor lnot B) land C)big) ;land; A
op=hint{use $;A;$ on other side of the rightmost $;land;$}
big(lnot (true lor B) land lnot C ;notequiv; lnot ((false lor lnot B) land C)big) ;land; A
op=hint{simplify}
big(false ;notequiv; lnot (lnot B land C)big) ;land; A
op=hint{simplify $;false notequiv P;$ to $;P;$; DeMorgan}
(B lor lnot C) ;land; A
endcalc$$
And now it is trivial to translate that back to boolean algebra notation...
answered Nov 12 '15 at 7:21
Marnix KloosterMarnix Klooster
4,20322146
I've never been a fan of the boolean algebra notation, so this is not a direct answer to your question, but more how I would do this simplification using propositional logic notation.$
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{ref}[1]{text{(#1)}}
newcommand{then}{Rightarrow}
newcommand{followsfrom}{Leftarrow}
newcommand{true}{text{true}}
newcommand{false}{text{false}}
$
Translated into that notation, you're asked to simplify $$
big(lnot (A lor B) land lnot C ;notequiv; lnot ((lnot A lor lnot B) land C)big) ;land; A
$$
We can do that as in the following calculation:
$$calc
big(lnot (A lor B) land lnot C ;notequiv; lnot ((lnot A lor lnot B) land C)big) ;land; A
op=hint{use $;A;$ on other side of the rightmost $;land;$}
big(lnot (true lor B) land lnot C ;notequiv; lnot ((false lor lnot B) land C)big) ;land; A
op=hint{simplify}
big(false ;notequiv; lnot (lnot B land C)big) ;land; A
op=hint{simplify $;false notequiv P;$ to $;P;$; DeMorgan}
(B lor lnot C) ;land; A
endcalc$$
And now it is trivial to translate that back to boolean algebra notation...
answered Nov 12 '15 at 7:21
Marnix KloosterMarnix Klooster
4,20322146
answered Nov 12 '15 at 7:21
Marnix KloosterMarnix Klooster
4,20322146
answered Nov 12 '15 at 7:21
Marnix KloosterMarnix Klooster
4,20322146
answered Nov 12 '15 at 7:21
Marnix KloosterMarnix Klooster
4,20322146
4,20322146
add a comment |
add a comment |
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What've you tried? What rules do you know and/or are you allowed to use?
– Marnix Klooster
Nov 11 '15 at 8:39