Suppose the logical expression $(((neg$ $P$) $leftrightarrow$ $Q$) $rightarrow$ $R$) $vee$ ($P$...
$begingroup$
Let $P$, $Q$, and $R$ be statement variables. Suppose the logical expression
$(((neg$$P$) $leftrightarrow$ $Q$) $rightarrow$ $R$) $vee$ ($P$ $leftrightarrow$ $R$)
is FALSE.
What are the possible truth values for $P$, $Q$, and $R$?
proof-writing
asked Jan 11 at 5:39
macymacy
62
$endgroup$
add a comment |
$begingroup$
Let $P$, $Q$, and $R$ be statement variables. Suppose the logical expression
$(((neg$$P$) $leftrightarrow$ $Q$) $rightarrow$ $R$) $vee$ ($P$ $leftrightarrow$ $R$)
is FALSE.
What are the possible truth values for $P$, $Q$, and $R$?
proof-writing
asked Jan 11 at 5:39
macymacy
62
$endgroup$
add a comment |
$begingroup$
Let $P$, $Q$, and $R$ be statement variables. Suppose the logical expression
$(((neg$$P$) $leftrightarrow$ $Q$) $rightarrow$ $R$) $vee$ ($P$ $leftrightarrow$ $R$)
is FALSE.
What are the possible truth values for $P$, $Q$, and $R$?
proof-writing
asked Jan 11 at 5:39
macymacy
62
$endgroup$
Let $P$, $Q$, and $R$ be statement variables. Suppose the logical expression
$(((neg$$P$) $leftrightarrow$ $Q$) $rightarrow$ $R$) $vee$ ($P$ $leftrightarrow$ $R$)
is FALSE.
What are the possible truth values for $P$, $Q$, and $R$?
proof-writing
proof-writing
asked Jan 11 at 5:39
macymacy
62
asked Jan 11 at 5:39
macymacy
62
asked Jan 11 at 5:39
macymacy
62
asked Jan 11 at 5:39
macymacy
62
asked Jan 11 at 5:39
macymacy
62
62
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Both $(neg Pleftrightarrow Q)to R$ and $Pleftrightarrow R$ must be false. The latter is false only when $P=1,R=0$ or $P=0,R=1$.
For the former to be false, $R=0,neg Pleftrightarrow Q=1$. This leaves you with $P=1,R=0$. Now try out $Q=0,1$ to see which one makes $neg Pleftrightarrow Q=1$.
You should get $P=1,R=0,Q=0$.
answered Jan 11 at 5:52
Shubham JohriShubham Johri
5,002717
$endgroup$
add a comment |
$begingroup$
The easiest (and also most tedious) route is to just make a truth table for that compound statement and see which values of $P$, $Q$ and $R$ make it false.
answered Jan 11 at 5:47
Stupid Questions IncStupid Questions Inc
7010
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
Both $(neg Pleftrightarrow Q)to R$ and $Pleftrightarrow R$ must be false. The latter is false only when $P=1,R=0$ or $P=0,R=1$.
For the former to be false, $R=0,neg Pleftrightarrow Q=1$. This leaves you with $P=1,R=0$. Now try out $Q=0,1$ to see which one makes $neg Pleftrightarrow Q=1$.
You should get $P=1,R=0,Q=0$.
answered Jan 11 at 5:52
Shubham JohriShubham Johri
5,002717
$endgroup$
add a comment |
$begingroup$
Both $(neg Pleftrightarrow Q)to R$ and $Pleftrightarrow R$ must be false. The latter is false only when $P=1,R=0$ or $P=0,R=1$.
For the former to be false, $R=0,neg Pleftrightarrow Q=1$. This leaves you with $P=1,R=0$. Now try out $Q=0,1$ to see which one makes $neg Pleftrightarrow Q=1$.
You should get $P=1,R=0,Q=0$.
answered Jan 11 at 5:52
Shubham JohriShubham Johri
5,002717
$endgroup$
add a comment |
$begingroup$
Both $(neg Pleftrightarrow Q)to R$ and $Pleftrightarrow R$ must be false. The latter is false only when $P=1,R=0$ or $P=0,R=1$.
For the former to be false, $R=0,neg Pleftrightarrow Q=1$. This leaves you with $P=1,R=0$. Now try out $Q=0,1$ to see which one makes $neg Pleftrightarrow Q=1$.
You should get $P=1,R=0,Q=0$.
answered Jan 11 at 5:52
Shubham JohriShubham Johri
5,002717
$endgroup$
Both $(neg Pleftrightarrow Q)to R$ and $Pleftrightarrow R$ must be false. The latter is false only when $P=1,R=0$ or $P=0,R=1$.
For the former to be false, $R=0,neg Pleftrightarrow Q=1$. This leaves you with $P=1,R=0$. Now try out $Q=0,1$ to see which one makes $neg Pleftrightarrow Q=1$.
You should get $P=1,R=0,Q=0$.
answered Jan 11 at 5:52
Shubham JohriShubham Johri
5,002717
answered Jan 11 at 5:52
Shubham JohriShubham Johri
5,002717
answered Jan 11 at 5:52
Shubham JohriShubham Johri
5,002717
answered Jan 11 at 5:52
Shubham JohriShubham Johri
5,002717
5,002717
add a comment |
add a comment |
$begingroup$
The easiest (and also most tedious) route is to just make a truth table for that compound statement and see which values of $P$, $Q$ and $R$ make it false.
answered Jan 11 at 5:47
Stupid Questions IncStupid Questions Inc
7010
$endgroup$
add a comment |
$begingroup$
The easiest (and also most tedious) route is to just make a truth table for that compound statement and see which values of $P$, $Q$ and $R$ make it false.
answered Jan 11 at 5:47
Stupid Questions IncStupid Questions Inc
7010
$endgroup$
add a comment |
$begingroup$
The easiest (and also most tedious) route is to just make a truth table for that compound statement and see which values of $P$, $Q$ and $R$ make it false.
answered Jan 11 at 5:47
Stupid Questions IncStupid Questions Inc
7010
$endgroup$
The easiest (and also most tedious) route is to just make a truth table for that compound statement and see which values of $P$, $Q$ and $R$ make it false.
answered Jan 11 at 5:47
Stupid Questions IncStupid Questions Inc
7010
answered Jan 11 at 5:47
Stupid Questions IncStupid Questions Inc
7010
answered Jan 11 at 5:47
Stupid Questions IncStupid Questions Inc
7010
answered Jan 11 at 5:47
Stupid Questions IncStupid Questions Inc
7010
7010
add a comment |
add a comment |
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