Mixing
Is it possible to calculate the logical expression A ^ B v C?
$begingroup$
In my theory notes, logical AND $land$ and logical OR $lor$ are equivalent in precedence. So then is the expression $A land B lor C$ valid, i.e. can I compute it, without parenthesis?
logic
edited Jan 26 at 16:22
Mauro ALLEGRANZA
67.4k449115
asked Jan 26 at 16:07
TMOTTMTMOTTM
1254
$endgroup$
add a comment |
$begingroup$
In my theory notes, logical AND $land$ and logical OR $lor$ are equivalent in precedence. So then is the expression $A land B lor C$ valid, i.e. can I compute it, without parenthesis?
logic
edited Jan 26 at 16:22
Mauro ALLEGRANZA
67.4k449115
asked Jan 26 at 16:07
TMOTTMTMOTTM
1254
$endgroup$
$begingroup$
The two ways of parsing it are certainly not equivalent. You should note that a more common convention is that it's $(Aland B)lor C$.
$endgroup$
– David C. Ullrich
Jan 26 at 17:45
$begingroup$
In our situation its explicitly not the convention.
$endgroup$
– TMOTTM
Jan 30 at 10:48
add a comment |
$begingroup$
In my theory notes, logical AND $land$ and logical OR $lor$ are equivalent in precedence. So then is the expression $A land B lor C$ valid, i.e. can I compute it, without parenthesis?
logic
edited Jan 26 at 16:22
Mauro ALLEGRANZA
67.4k449115
asked Jan 26 at 16:07
TMOTTMTMOTTM
1254
$endgroup$
In my theory notes, logical AND $land$ and logical OR $lor$ are equivalent in precedence. So then is the expression $A land B lor C$ valid, i.e. can I compute it, without parenthesis?
logic
logic
edited Jan 26 at 16:22
Mauro ALLEGRANZA
67.4k449115
asked Jan 26 at 16:07
TMOTTMTMOTTM
1254
edited Jan 26 at 16:22
Mauro ALLEGRANZA
67.4k449115
asked Jan 26 at 16:07
TMOTTMTMOTTM
1254
edited Jan 26 at 16:22
Mauro ALLEGRANZA
67.4k449115
edited Jan 26 at 16:22
Mauro ALLEGRANZA
67.4k449115
edited Jan 26 at 16:22
Mauro ALLEGRANZA
67.4k449115
67.4k449115
asked Jan 26 at 16:07
TMOTTMTMOTTM
1254
asked Jan 26 at 16:07
TMOTTMTMOTTM
1254
asked Jan 26 at 16:07
TMOTTMTMOTTM
1254
1254
$begingroup$
The two ways of parsing it are certainly not equivalent. You should note that a more common convention is that it's $(Aland B)lor C$.
$endgroup$
– David C. Ullrich
Jan 26 at 17:45
$begingroup$
In our situation its explicitly not the convention.
$endgroup$
– TMOTTM
Jan 30 at 10:48
add a comment |
$begingroup$
The two ways of parsing it are certainly not equivalent. You should note that a more common convention is that it's $(Aland B)lor C$.
$endgroup$
– David C. Ullrich
Jan 26 at 17:45
$begingroup$
In our situation its explicitly not the convention.
$endgroup$
– TMOTTM
Jan 30 at 10:48
$begingroup$
The two ways of parsing it are certainly not equivalent. You should note that a more common convention is that it's $(Aland B)lor C$.
$endgroup$
– David C. Ullrich
Jan 26 at 17:45
$begingroup$
The two ways of parsing it are certainly not equivalent. You should note that a more common convention is that it's $(Aland B)lor C$.
$endgroup$
– David C. Ullrich
Jan 26 at 17:45
$begingroup$
In our situation its explicitly not the convention.
$endgroup$
– TMOTTM
Jan 30 at 10:48
$begingroup$
In our situation its explicitly not the convention.
$endgroup$
– TMOTTM
Jan 30 at 10:48
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
No, its not. Consider $A$ false and $B,C$ true. Then $fwedge (tvee t) = f$ and $(fwedge t)vee t = t$.
answered Jan 26 at 16:11
WuestenfuxWuestenfux
5,2931513
$endgroup$
add a comment |
$begingroup$
No, without parentheses or a convention about which connective is the stronger connective, the expression is ambiguous, since $(A wedge B) vee C$ and $A wedge (B vee C)$ are not equivalent.
answered Jan 26 at 16:08
kccukccu
10.6k11229
$endgroup$
$begingroup$
As I was thinking. In fact our convention is that they are of same connectivity.
$endgroup$
– TMOTTM
Jan 26 at 16:10
$begingroup$
Right, there convention is that they are equally strong: proofwiki.org/wiki/Definition:Binding_Priority.
$endgroup$
– kccu
Jan 26 at 16:14
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
No, its not. Consider $A$ false and $B,C$ true. Then $fwedge (tvee t) = f$ and $(fwedge t)vee t = t$.
answered Jan 26 at 16:11
WuestenfuxWuestenfux
5,2931513
$endgroup$
add a comment |
$begingroup$
No, its not. Consider $A$ false and $B,C$ true. Then $fwedge (tvee t) = f$ and $(fwedge t)vee t = t$.
answered Jan 26 at 16:11
WuestenfuxWuestenfux
5,2931513
$endgroup$
add a comment |
$begingroup$
No, its not. Consider $A$ false and $B,C$ true. Then $fwedge (tvee t) = f$ and $(fwedge t)vee t = t$.
answered Jan 26 at 16:11
WuestenfuxWuestenfux
5,2931513
$endgroup$
No, its not. Consider $A$ false and $B,C$ true. Then $fwedge (tvee t) = f$ and $(fwedge t)vee t = t$.
answered Jan 26 at 16:11
WuestenfuxWuestenfux
5,2931513
answered Jan 26 at 16:11
WuestenfuxWuestenfux
5,2931513
answered Jan 26 at 16:11
WuestenfuxWuestenfux
5,2931513
answered Jan 26 at 16:11
WuestenfuxWuestenfux
5,2931513
5,2931513
add a comment |
add a comment |
$begingroup$
No, without parentheses or a convention about which connective is the stronger connective, the expression is ambiguous, since $(A wedge B) vee C$ and $A wedge (B vee C)$ are not equivalent.
answered Jan 26 at 16:08
kccukccu
10.6k11229
$endgroup$
$begingroup$
As I was thinking. In fact our convention is that they are of same connectivity.
$endgroup$
– TMOTTM
Jan 26 at 16:10
$begingroup$
Right, there convention is that they are equally strong: proofwiki.org/wiki/Definition:Binding_Priority.
$endgroup$
– kccu
Jan 26 at 16:14
add a comment |
$begingroup$
No, without parentheses or a convention about which connective is the stronger connective, the expression is ambiguous, since $(A wedge B) vee C$ and $A wedge (B vee C)$ are not equivalent.
answered Jan 26 at 16:08
kccukccu
10.6k11229
$endgroup$
$begingroup$
As I was thinking. In fact our convention is that they are of same connectivity.
$endgroup$
– TMOTTM
Jan 26 at 16:10
$begingroup$
Right, there convention is that they are equally strong: proofwiki.org/wiki/Definition:Binding_Priority.
$endgroup$
– kccu
Jan 26 at 16:14
add a comment |
$begingroup$
No, without parentheses or a convention about which connective is the stronger connective, the expression is ambiguous, since $(A wedge B) vee C$ and $A wedge (B vee C)$ are not equivalent.
answered Jan 26 at 16:08
kccukccu
10.6k11229
$endgroup$
No, without parentheses or a convention about which connective is the stronger connective, the expression is ambiguous, since $(A wedge B) vee C$ and $A wedge (B vee C)$ are not equivalent.
answered Jan 26 at 16:08
kccukccu
10.6k11229
answered Jan 26 at 16:08
kccukccu
10.6k11229
answered Jan 26 at 16:08
kccukccu
10.6k11229
answered Jan 26 at 16:08
kccukccu
10.6k11229
10.6k11229
$begingroup$
As I was thinking. In fact our convention is that they are of same connectivity.
$endgroup$
– TMOTTM
Jan 26 at 16:10
$begingroup$
Right, there convention is that they are equally strong: proofwiki.org/wiki/Definition:Binding_Priority.
$endgroup$
– kccu
Jan 26 at 16:14
add a comment |
$begingroup$
As I was thinking. In fact our convention is that they are of same connectivity.
$endgroup$
– TMOTTM
Jan 26 at 16:10
$begingroup$
Right, there convention is that they are equally strong: proofwiki.org/wiki/Definition:Binding_Priority.
$endgroup$
– kccu
Jan 26 at 16:14
$begingroup$
As I was thinking. In fact our convention is that they are of same connectivity.
$endgroup$
– TMOTTM
Jan 26 at 16:10
$begingroup$
As I was thinking. In fact our convention is that they are of same connectivity.
$endgroup$
– TMOTTM
Jan 26 at 16:10
$begingroup$
Right, there convention is that they are equally strong: proofwiki.org/wiki/Definition:Binding_Priority.
$endgroup$
– kccu
Jan 26 at 16:14
$begingroup$
Right, there convention is that they are equally strong: proofwiki.org/wiki/Definition:Binding_Priority.
$endgroup$
– kccu
Jan 26 at 16:14
add a comment |
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$begingroup$
The two ways of parsing it are certainly not equivalent. You should note that a more common convention is that it's $(Aland B)lor C$.
$endgroup$
– David C. Ullrich
Jan 26 at 17:45
$begingroup$
In our situation its explicitly not the convention.
$endgroup$
– TMOTTM
Jan 30 at 10:48