Finding operand from logical operation
$begingroup$
If the truth value R can be determined with this logic:
$$
O_1 oplus O_2 oplus C_text{in} = R
$$
And I am given the values of $O_1, O_2, text{and } R$, what operation can I perform to get the value of $C_text{in}$?
Edit: After staring at a truth table for a while, am I right in thinking that:
$
O_1 oplus O_2 oplus R = C_text{in}
$?
logic propositional-calculus
asked Dec 8 '18 at 1:30
NoahNoah
52
$endgroup$
add a comment |
$begingroup$
If the truth value R can be determined with this logic:
$$
O_1 oplus O_2 oplus C_text{in} = R
$$
And I am given the values of $O_1, O_2, text{and } R$, what operation can I perform to get the value of $C_text{in}$?
Edit: After staring at a truth table for a while, am I right in thinking that:
$
O_1 oplus O_2 oplus R = C_text{in}
$?
logic propositional-calculus
asked Dec 8 '18 at 1:30
NoahNoah
52
$endgroup$
add a comment |
$begingroup$
If the truth value R can be determined with this logic:
$$
O_1 oplus O_2 oplus C_text{in} = R
$$
And I am given the values of $O_1, O_2, text{and } R$, what operation can I perform to get the value of $C_text{in}$?
Edit: After staring at a truth table for a while, am I right in thinking that:
$
O_1 oplus O_2 oplus R = C_text{in}
$?
logic propositional-calculus
asked Dec 8 '18 at 1:30
NoahNoah
52
$endgroup$
If the truth value R can be determined with this logic:
$$
O_1 oplus O_2 oplus C_text{in} = R
$$
And I am given the values of $O_1, O_2, text{and } R$, what operation can I perform to get the value of $C_text{in}$?
Edit: After staring at a truth table for a while, am I right in thinking that:
$
O_1 oplus O_2 oplus R = C_text{in}
$?
logic propositional-calculus
logic propositional-calculus
asked Dec 8 '18 at 1:30
NoahNoah
52
asked Dec 8 '18 at 1:30
NoahNoah
52
edited Dec 8 '18 at 1:45
Noah
asked Dec 8 '18 at 1:30
NoahNoah
52
asked Dec 8 '18 at 1:30
NoahNoah
52
asked Dec 8 '18 at 1:30
NoahNoah
52
52
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes.
You can derive that from the following three rules for $oplus$:
- $x oplus (y oplus z) = (x oplus y) oplus z$
- $(x oplus x) oplus y = y$
- $x oplus y = y oplus xthinspace$
Just $oplus (C_{in} oplus R)$ on both sides.
Staring at a truth table is also a pretty nice way though.
answered Dec 8 '18 at 2:01
Y.DingY.Ding
986
$endgroup$
1
$begingroup$
Ah, right, three!
$endgroup$
– Y.Ding
Dec 9 '18 at 7:21
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes.
You can derive that from the following three rules for $oplus$:
- $x oplus (y oplus z) = (x oplus y) oplus z$
- $(x oplus x) oplus y = y$
- $x oplus y = y oplus xthinspace$
Just $oplus (C_{in} oplus R)$ on both sides.
Staring at a truth table is also a pretty nice way though.
answered Dec 8 '18 at 2:01
Y.DingY.Ding
986
$endgroup$
1
$begingroup$
Ah, right, three!
$endgroup$
– Y.Ding
Dec 9 '18 at 7:21
add a comment |
$begingroup$
Yes.
You can derive that from the following three rules for $oplus$:
- $x oplus (y oplus z) = (x oplus y) oplus z$
- $(x oplus x) oplus y = y$
- $x oplus y = y oplus xthinspace$
Just $oplus (C_{in} oplus R)$ on both sides.
Staring at a truth table is also a pretty nice way though.
answered Dec 8 '18 at 2:01
Y.DingY.Ding
986
$endgroup$
1
$begingroup$
Ah, right, three!
$endgroup$
– Y.Ding
Dec 9 '18 at 7:21
add a comment |
$begingroup$
Yes.
You can derive that from the following three rules for $oplus$:
- $x oplus (y oplus z) = (x oplus y) oplus z$
- $(x oplus x) oplus y = y$
- $x oplus y = y oplus xthinspace$
Just $oplus (C_{in} oplus R)$ on both sides.
Staring at a truth table is also a pretty nice way though.
answered Dec 8 '18 at 2:01
Y.DingY.Ding
986
$endgroup$
Yes.
You can derive that from the following three rules for $oplus$:
- $x oplus (y oplus z) = (x oplus y) oplus z$
- $(x oplus x) oplus y = y$
- $x oplus y = y oplus xthinspace$
Just $oplus (C_{in} oplus R)$ on both sides.
Staring at a truth table is also a pretty nice way though.
answered Dec 8 '18 at 2:01
Y.DingY.Ding
986
edited Dec 9 '18 at 7:17
answered Dec 8 '18 at 2:01
Y.DingY.Ding
986
answered Dec 8 '18 at 2:01
Y.DingY.Ding
986
answered Dec 8 '18 at 2:01
Y.DingY.Ding
986
986
1
$begingroup$
Ah, right, three!
$endgroup$
– Y.Ding
Dec 9 '18 at 7:21
add a comment |
1
$begingroup$
Ah, right, three!
$endgroup$
– Y.Ding
Dec 9 '18 at 7:21
1
1
$begingroup$
Ah, right, three!
$endgroup$
– Y.Ding
Dec 9 '18 at 7:21
$begingroup$
Ah, right, three!
$endgroup$
– Y.Ding
Dec 9 '18 at 7:21
add a comment |
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