What's the difference between biconditional iff and logical equivalence?
$begingroup$
I am confused about the difference between ↔ (biconditional iff) and ≡ (logical equivalence). For instance, p→q can be rewritten as ∼p∨q. Would it be correct to say p→q↔∼p∨q or p→q≡∼p∨q?
Secondly, is ⇔ another symbol for ≡?
Finally, what's the difference between → and ⇒?
logic propositional-calculus
edited Sep 18 '17 at 6:57
Mauro ALLEGRANZA
65.6k449113
asked Sep 17 '17 at 0:24
doomblahdoomblah
63
$endgroup$
add a comment |
$begingroup$
I am confused about the difference between ↔ (biconditional iff) and ≡ (logical equivalence). For instance, p→q can be rewritten as ∼p∨q. Would it be correct to say p→q↔∼p∨q or p→q≡∼p∨q?
Secondly, is ⇔ another symbol for ≡?
Finally, what's the difference between → and ⇒?
logic propositional-calculus
edited Sep 18 '17 at 6:57
Mauro ALLEGRANZA
65.6k449113
asked Sep 17 '17 at 0:24
doomblahdoomblah
63
$endgroup$
2
$begingroup$
You should look in that book to see what the symbols mean. For example, it could be that $rightarrow$ combines two wffs into a new wff, while $Rightarrow$ is a relation between wffs.
$endgroup$
– GEdgar
Sep 17 '17 at 0:31
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The issue is not with the symbols but with the concepts. We have a connective: the biconditional that can produce a "complex" sentence (or formula) from simpler ones: $p leftrightarrow q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:54
$begingroup$
And we have the relation of logical (or semantical) equivalence between formulas. Logical equivalence is different from the biconditional, although the two concepts are closely related; in a nutshell: $p leftrightarrow q$ is a tautology iff $p$ is logically equivalent to $q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:57
add a comment |
$begingroup$
I am confused about the difference between ↔ (biconditional iff) and ≡ (logical equivalence). For instance, p→q can be rewritten as ∼p∨q. Would it be correct to say p→q↔∼p∨q or p→q≡∼p∨q?
Secondly, is ⇔ another symbol for ≡?
Finally, what's the difference between → and ⇒?
logic propositional-calculus
edited Sep 18 '17 at 6:57
Mauro ALLEGRANZA
65.6k449113
asked Sep 17 '17 at 0:24
doomblahdoomblah
63
$endgroup$
I am confused about the difference between ↔ (biconditional iff) and ≡ (logical equivalence). For instance, p→q can be rewritten as ∼p∨q. Would it be correct to say p→q↔∼p∨q or p→q≡∼p∨q?
Secondly, is ⇔ another symbol for ≡?
Finally, what's the difference between → and ⇒?
logic propositional-calculus
logic propositional-calculus
edited Sep 18 '17 at 6:57
Mauro ALLEGRANZA
65.6k449113
asked Sep 17 '17 at 0:24
doomblahdoomblah
63
edited Sep 18 '17 at 6:57
Mauro ALLEGRANZA
65.6k449113
asked Sep 17 '17 at 0:24
doomblahdoomblah
63
edited Sep 18 '17 at 6:57
Mauro ALLEGRANZA
65.6k449113
edited Sep 18 '17 at 6:57
Mauro ALLEGRANZA
65.6k449113
edited Sep 18 '17 at 6:57
Mauro ALLEGRANZA
65.6k449113
65.6k449113
asked Sep 17 '17 at 0:24
doomblahdoomblah
63
asked Sep 17 '17 at 0:24
doomblahdoomblah
63
asked Sep 17 '17 at 0:24
doomblahdoomblah
63
63
2
$begingroup$
You should look in that book to see what the symbols mean. For example, it could be that $rightarrow$ combines two wffs into a new wff, while $Rightarrow$ is a relation between wffs.
$endgroup$
– GEdgar
Sep 17 '17 at 0:31
$begingroup$
The issue is not with the symbols but with the concepts. We have a connective: the biconditional that can produce a "complex" sentence (or formula) from simpler ones: $p leftrightarrow q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:54
$begingroup$
And we have the relation of logical (or semantical) equivalence between formulas. Logical equivalence is different from the biconditional, although the two concepts are closely related; in a nutshell: $p leftrightarrow q$ is a tautology iff $p$ is logically equivalent to $q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:57
add a comment |
2
$begingroup$
You should look in that book to see what the symbols mean. For example, it could be that $rightarrow$ combines two wffs into a new wff, while $Rightarrow$ is a relation between wffs.
$endgroup$
– GEdgar
Sep 17 '17 at 0:31
$begingroup$
The issue is not with the symbols but with the concepts. We have a connective: the biconditional that can produce a "complex" sentence (or formula) from simpler ones: $p leftrightarrow q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:54
$begingroup$
And we have the relation of logical (or semantical) equivalence between formulas. Logical equivalence is different from the biconditional, although the two concepts are closely related; in a nutshell: $p leftrightarrow q$ is a tautology iff $p$ is logically equivalent to $q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:57
2
2
$begingroup$
You should look in that book to see what the symbols mean. For example, it could be that $rightarrow$ combines two wffs into a new wff, while $Rightarrow$ is a relation between wffs.
$endgroup$
– GEdgar
Sep 17 '17 at 0:31
$begingroup$
You should look in that book to see what the symbols mean. For example, it could be that $rightarrow$ combines two wffs into a new wff, while $Rightarrow$ is a relation between wffs.
$endgroup$
– GEdgar
Sep 17 '17 at 0:31
$begingroup$
The issue is not with the symbols but with the concepts. We have a connective: the biconditional that can produce a "complex" sentence (or formula) from simpler ones: $p leftrightarrow q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:54
$begingroup$
The issue is not with the symbols but with the concepts. We have a connective: the biconditional that can produce a "complex" sentence (or formula) from simpler ones: $p leftrightarrow q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:54
$begingroup$
And we have the relation of logical (or semantical) equivalence between formulas. Logical equivalence is different from the biconditional, although the two concepts are closely related; in a nutshell: $p leftrightarrow q$ is a tautology iff $p$ is logically equivalent to $q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:57
$begingroup$
And we have the relation of logical (or semantical) equivalence between formulas. Logical equivalence is different from the biconditional, although the two concepts are closely related; in a nutshell: $p leftrightarrow q$ is a tautology iff $p$ is logically equivalent to $q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In short, $P leftrightarrow Q$ is statement that could be either true or false. $P equiv Q$ means that $P leftrightarrow Q$ is always a true biconditional (so, $P$ and $Q$ have the same truth value no matter what).
So, one could say that $neg (P vee Q) equiv neg P wedge neg Q$ (DeMorgan's) but you typically wouldn't write $neg (P vee Q) leftrightarrow (neg P wedge neg Q)$.
The arrow $Rightarrow$ usually is slang for "implies" but different people use it differently. The arrow $Leftrightarrow$ is usually treated the same way as $leftrightarrow$.
answered Sep 17 '17 at 0:48
RandallRandall
9,71111230
$endgroup$
add a comment |
$begingroup$
In case you can use a somewhat philosophical explanation: $leftrightarrow$ is a logical operator within statements, while $equiv$ serves to state an equivalence between statements and thus may be thought of as meta-logical.
As Randall explained, $P leftrightarrow Q$ is a statement $-$ one statement and a logical statement. The $leftrightarrow$ will cause it to be true under certain truth value distributions for $P$ and $Q$. The same applies to $neg(P vee Q) leftrightarrow (neg P wedge neg Q)$.
For $neg(P vee Q) equiv (neg P wedge neg Q)$ however, you compare two truth tables, the one of $neg(P vee Q)$ and $(neg P wedge neg Q)$, two distinct statements. If and only if both are true exclusively for the same truth value distributions, the equivalence applies and so the meta-logical statement $neg(P vee Q) equiv (neg P wedge neg Q)$ is true.
answered Sep 17 '17 at 4:51
RauteRaute
2817
$endgroup$
add a comment |
$begingroup$
Consider this analogy. You wouldn't say the following:
Prove that 3 + 5.
You might instead say this:
Prove that 3 = 5.
Similarly, following doesn't make sense:
Prove that $p leftrightarrow q$.
Instead one of these would be correct:
Prove that $p leftrightarrow q$ is always true.
or
Prove that $p equiv q$.
answered Apr 22 '18 at 3:56
Silap AliyevSilap Aliyev
1136
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add a comment |
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3 Answers
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3 Answers
3
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$begingroup$
In short, $P leftrightarrow Q$ is statement that could be either true or false. $P equiv Q$ means that $P leftrightarrow Q$ is always a true biconditional (so, $P$ and $Q$ have the same truth value no matter what).
So, one could say that $neg (P vee Q) equiv neg P wedge neg Q$ (DeMorgan's) but you typically wouldn't write $neg (P vee Q) leftrightarrow (neg P wedge neg Q)$.
The arrow $Rightarrow$ usually is slang for "implies" but different people use it differently. The arrow $Leftrightarrow$ is usually treated the same way as $leftrightarrow$.
answered Sep 17 '17 at 0:48
RandallRandall
9,71111230
$endgroup$
add a comment |
$begingroup$
In short, $P leftrightarrow Q$ is statement that could be either true or false. $P equiv Q$ means that $P leftrightarrow Q$ is always a true biconditional (so, $P$ and $Q$ have the same truth value no matter what).
So, one could say that $neg (P vee Q) equiv neg P wedge neg Q$ (DeMorgan's) but you typically wouldn't write $neg (P vee Q) leftrightarrow (neg P wedge neg Q)$.
The arrow $Rightarrow$ usually is slang for "implies" but different people use it differently. The arrow $Leftrightarrow$ is usually treated the same way as $leftrightarrow$.
answered Sep 17 '17 at 0:48
RandallRandall
9,71111230
$endgroup$
add a comment |
$begingroup$
In short, $P leftrightarrow Q$ is statement that could be either true or false. $P equiv Q$ means that $P leftrightarrow Q$ is always a true biconditional (so, $P$ and $Q$ have the same truth value no matter what).
So, one could say that $neg (P vee Q) equiv neg P wedge neg Q$ (DeMorgan's) but you typically wouldn't write $neg (P vee Q) leftrightarrow (neg P wedge neg Q)$.
The arrow $Rightarrow$ usually is slang for "implies" but different people use it differently. The arrow $Leftrightarrow$ is usually treated the same way as $leftrightarrow$.
answered Sep 17 '17 at 0:48
RandallRandall
9,71111230
$endgroup$
In short, $P leftrightarrow Q$ is statement that could be either true or false. $P equiv Q$ means that $P leftrightarrow Q$ is always a true biconditional (so, $P$ and $Q$ have the same truth value no matter what).
So, one could say that $neg (P vee Q) equiv neg P wedge neg Q$ (DeMorgan's) but you typically wouldn't write $neg (P vee Q) leftrightarrow (neg P wedge neg Q)$.
The arrow $Rightarrow$ usually is slang for "implies" but different people use it differently. The arrow $Leftrightarrow$ is usually treated the same way as $leftrightarrow$.
answered Sep 17 '17 at 0:48
RandallRandall
9,71111230
answered Sep 17 '17 at 0:48
RandallRandall
9,71111230
answered Sep 17 '17 at 0:48
RandallRandall
9,71111230
answered Sep 17 '17 at 0:48
RandallRandall
9,71111230
9,71111230
add a comment |
add a comment |
$begingroup$
In case you can use a somewhat philosophical explanation: $leftrightarrow$ is a logical operator within statements, while $equiv$ serves to state an equivalence between statements and thus may be thought of as meta-logical.
As Randall explained, $P leftrightarrow Q$ is a statement $-$ one statement and a logical statement. The $leftrightarrow$ will cause it to be true under certain truth value distributions for $P$ and $Q$. The same applies to $neg(P vee Q) leftrightarrow (neg P wedge neg Q)$.
For $neg(P vee Q) equiv (neg P wedge neg Q)$ however, you compare two truth tables, the one of $neg(P vee Q)$ and $(neg P wedge neg Q)$, two distinct statements. If and only if both are true exclusively for the same truth value distributions, the equivalence applies and so the meta-logical statement $neg(P vee Q) equiv (neg P wedge neg Q)$ is true.
answered Sep 17 '17 at 4:51
RauteRaute
2817
$endgroup$
add a comment |
$begingroup$
In case you can use a somewhat philosophical explanation: $leftrightarrow$ is a logical operator within statements, while $equiv$ serves to state an equivalence between statements and thus may be thought of as meta-logical.
As Randall explained, $P leftrightarrow Q$ is a statement $-$ one statement and a logical statement. The $leftrightarrow$ will cause it to be true under certain truth value distributions for $P$ and $Q$. The same applies to $neg(P vee Q) leftrightarrow (neg P wedge neg Q)$.
For $neg(P vee Q) equiv (neg P wedge neg Q)$ however, you compare two truth tables, the one of $neg(P vee Q)$ and $(neg P wedge neg Q)$, two distinct statements. If and only if both are true exclusively for the same truth value distributions, the equivalence applies and so the meta-logical statement $neg(P vee Q) equiv (neg P wedge neg Q)$ is true.
answered Sep 17 '17 at 4:51
RauteRaute
2817
$endgroup$
add a comment |
$begingroup$
In case you can use a somewhat philosophical explanation: $leftrightarrow$ is a logical operator within statements, while $equiv$ serves to state an equivalence between statements and thus may be thought of as meta-logical.
As Randall explained, $P leftrightarrow Q$ is a statement $-$ one statement and a logical statement. The $leftrightarrow$ will cause it to be true under certain truth value distributions for $P$ and $Q$. The same applies to $neg(P vee Q) leftrightarrow (neg P wedge neg Q)$.
For $neg(P vee Q) equiv (neg P wedge neg Q)$ however, you compare two truth tables, the one of $neg(P vee Q)$ and $(neg P wedge neg Q)$, two distinct statements. If and only if both are true exclusively for the same truth value distributions, the equivalence applies and so the meta-logical statement $neg(P vee Q) equiv (neg P wedge neg Q)$ is true.
answered Sep 17 '17 at 4:51
RauteRaute
2817
$endgroup$
In case you can use a somewhat philosophical explanation: $leftrightarrow$ is a logical operator within statements, while $equiv$ serves to state an equivalence between statements and thus may be thought of as meta-logical.
As Randall explained, $P leftrightarrow Q$ is a statement $-$ one statement and a logical statement. The $leftrightarrow$ will cause it to be true under certain truth value distributions for $P$ and $Q$. The same applies to $neg(P vee Q) leftrightarrow (neg P wedge neg Q)$.
For $neg(P vee Q) equiv (neg P wedge neg Q)$ however, you compare two truth tables, the one of $neg(P vee Q)$ and $(neg P wedge neg Q)$, two distinct statements. If and only if both are true exclusively for the same truth value distributions, the equivalence applies and so the meta-logical statement $neg(P vee Q) equiv (neg P wedge neg Q)$ is true.
answered Sep 17 '17 at 4:51
RauteRaute
2817
answered Sep 17 '17 at 4:51
RauteRaute
2817
answered Sep 17 '17 at 4:51
RauteRaute
2817
answered Sep 17 '17 at 4:51
RauteRaute
2817
2817
add a comment |
add a comment |
$begingroup$
Consider this analogy. You wouldn't say the following:
Prove that 3 + 5.
You might instead say this:
Prove that 3 = 5.
Similarly, following doesn't make sense:
Prove that $p leftrightarrow q$.
Instead one of these would be correct:
Prove that $p leftrightarrow q$ is always true.
or
Prove that $p equiv q$.
answered Apr 22 '18 at 3:56
Silap AliyevSilap Aliyev
1136
$endgroup$
add a comment |
$begingroup$
Consider this analogy. You wouldn't say the following:
Prove that 3 + 5.
You might instead say this:
Prove that 3 = 5.
Similarly, following doesn't make sense:
Prove that $p leftrightarrow q$.
Instead one of these would be correct:
Prove that $p leftrightarrow q$ is always true.
or
Prove that $p equiv q$.
answered Apr 22 '18 at 3:56
Silap AliyevSilap Aliyev
1136
$endgroup$
add a comment |
$begingroup$
Consider this analogy. You wouldn't say the following:
Prove that 3 + 5.
You might instead say this:
Prove that 3 = 5.
Similarly, following doesn't make sense:
Prove that $p leftrightarrow q$.
Instead one of these would be correct:
Prove that $p leftrightarrow q$ is always true.
or
Prove that $p equiv q$.
answered Apr 22 '18 at 3:56
Silap AliyevSilap Aliyev
1136
$endgroup$
Consider this analogy. You wouldn't say the following:
Prove that 3 + 5.
You might instead say this:
Prove that 3 = 5.
Similarly, following doesn't make sense:
Prove that $p leftrightarrow q$.
Instead one of these would be correct:
Prove that $p leftrightarrow q$ is always true.
or
Prove that $p equiv q$.
answered Apr 22 '18 at 3:56
Silap AliyevSilap Aliyev
1136
edited Dec 5 '18 at 7:26
answered Apr 22 '18 at 3:56
Silap AliyevSilap Aliyev
1136
answered Apr 22 '18 at 3:56
Silap AliyevSilap Aliyev
1136
answered Apr 22 '18 at 3:56
Silap AliyevSilap Aliyev
1136
1136
add a comment |
add a comment |
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2
$begingroup$
You should look in that book to see what the symbols mean. For example, it could be that $rightarrow$ combines two wffs into a new wff, while $Rightarrow$ is a relation between wffs.
$endgroup$
– GEdgar
Sep 17 '17 at 0:31
$begingroup$
The issue is not with the symbols but with the concepts. We have a connective: the biconditional that can produce a "complex" sentence (or formula) from simpler ones: $p leftrightarrow q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:54
$begingroup$
And we have the relation of logical (or semantical) equivalence between formulas. Logical equivalence is different from the biconditional, although the two concepts are closely related; in a nutshell: $p leftrightarrow q$ is a tautology iff $p$ is logically equivalent to $q$.
$endgroup$
– Mauro ALLEGRANZA
Sep 18 '17 at 6:57