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Is This Logic Statement valid?
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$neg(Arightarrow B)equiv neg(neg Avee B)equiv (Awedge neg B)$
logic propositional-calculus
edited Aug 2 at 14:37
zzuussee
1,142419
asked Aug 2 at 14:05
user6394019
30311
add a comment |Â
up vote
2
down vote
favorite
$neg(Arightarrow B)equiv neg(neg Avee B)equiv (Awedge neg B)$
logic propositional-calculus
edited Aug 2 at 14:37
zzuussee
1,142419
asked Aug 2 at 14:05
user6394019
30311
1
I just checked this and I could not find any mistake so I guess yes.
– mrtaurho
Aug 2 at 14:08
1
It is not really proof verification though.
– zzuussee
Aug 2 at 14:19
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$neg(Arightarrow B)equiv neg(neg Avee B)equiv (Awedge neg B)$
logic propositional-calculus
edited Aug 2 at 14:37
zzuussee
1,142419
asked Aug 2 at 14:05
user6394019
30311
$neg(Arightarrow B)equiv neg(neg Avee B)equiv (Awedge neg B)$
logic propositional-calculus
edited Aug 2 at 14:37
zzuussee
1,142419
asked Aug 2 at 14:05
user6394019
30311
edited Aug 2 at 14:37
zzuussee
1,142419
edited Aug 2 at 14:37
zzuussee
1,142419
edited Aug 2 at 14:37
zzuussee
1,142419
1,142419
asked Aug 2 at 14:05
user6394019
30311
asked Aug 2 at 14:05
user6394019
30311
asked Aug 2 at 14:05
user6394019
30311
30311
1
I just checked this and I could not find any mistake so I guess yes.
– mrtaurho
Aug 2 at 14:08
1
It is not really proof verification though.
– zzuussee
Aug 2 at 14:19
add a comment |Â
1
I just checked this and I could not find any mistake so I guess yes.
– mrtaurho
Aug 2 at 14:08
1
It is not really proof verification though.
– zzuussee
Aug 2 at 14:19
1
1
I just checked this and I could not find any mistake so I guess yes.
– mrtaurho
Aug 2 at 14:08
I just checked this and I could not find any mistake so I guess yes.
– mrtaurho
Aug 2 at 14:08
1
1
It is not really proof verification though.
– zzuussee
Aug 2 at 14:19
It is not really proof verification though.
– zzuussee
Aug 2 at 14:19
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
First equivalence is the defnition of $to$, and the second one is De Morgan's law plus double negation. Looks fine to me.
answered Aug 2 at 14:11
Arthur
98.2k792174
add a comment |Â
up vote
0
down vote
You can also observe this easily via the classical method of proof tables in propositional calculus:
beginarray
hline
A& B & neg(Arightarrow B) & neg(neg Alor B) & Alandneg B \ hline
0& 0& 0& 0& 0\ hline
0& 1& 0& 0& 0\ hline
1& 0& 1& 1& 1\ hline
1& 1& 0& 0& 0\ hline
endarray
by using the classical definitions for the elementary operators:
- Negation: beginarray hline A& neg A \ hline 0& 1\ hline 1& 0\
hline endarray - Conjunction: beginarray
hline
A& B & Aland B\ hline
0& 0& 0\ hline
0& 1& 0\ hline
1& 0& 0\ hline
1& 1& 1\ hline
endarray - Disjunction: beginarray
hline
A& B & Alor B\ hline
0& 0& 0\ hline
0& 1& 1\ hline
1& 0& 1\ hline
1& 1& 1\ hline
endarray - Implication: beginarray
hline
A& B & Arightarrow B\ hline
0& 0& 1\ hline
0& 1& 1\ hline
1& 0& 0\ hline
1& 1& 1\ hline
endarray
Thus, as $Cequiv D$ is valid if and only if the have the same truth value under every interpretation, we get the result.
answered Aug 2 at 14:19
zzuussee
1,142419
add a comment |Â
up vote
0
down vote
Yes or no, depending on exactly what you mean.
When people write $Pequiv Qequiv R$ in an informal context they usually mean that $P$ is equivalent to $Q$ and also $Q$ is equivalent to $R$. If that's what you meant then the answer to your question is yes, all three of those formulas are equivalent (if this is not clear to you you really haven't been paying attention in class - it's trivial to verify from truth tables).
But writing $Pequiv Qequiv R$ with the intended meaning as above is really bad notation, especially if, as here, $P$, $Q$ and $R$ are wffs in some formal system. Because then it looks as though $Pequiv Qequiv R$ is meant as a single formula in that system; if so it's not well formed, needing to be parenthesized as $(Pequiv Q)equiv R$ or $Pequiv (Qequiv R)$, neither of which "means" $P$ is equivalent to $Q$ and $Q$ is equivalent to $R$.
So: In the title you refer to "[t]his logical statement". If you mean to be asking about $neg(Arightarrow B)equiv neg(neg Avee B)equiv (Awedge neg B)$ as a wff in propositional logic then it is certainly not valid; neither of the interpretations $(neg(Arightarrow B)equiv neg(neg Avee B))equiv (Awedge neg B)$ and $neg(Arightarrow B)equiv (neg(neg Avee B)equiv (Awedge neg B))$ is valid, since they both say a tautology is equivalent to a non-tautology.
Moral This issue is why people write "The following are equivalent: $P$, $Q$, $R$" instead of writing $Pequiv Qequiv R$.
answered Aug 2 at 15:56
David C. Ullrich
53.8k33481
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
First equivalence is the defnition of $to$, and the second one is De Morgan's law plus double negation. Looks fine to me.
answered Aug 2 at 14:11
Arthur
98.2k792174
add a comment |Â
up vote
5
down vote
First equivalence is the defnition of $to$, and the second one is De Morgan's law plus double negation. Looks fine to me.
answered Aug 2 at 14:11
Arthur
98.2k792174
add a comment |Â
up vote
5
down vote
up vote
5
down vote
First equivalence is the defnition of $to$, and the second one is De Morgan's law plus double negation. Looks fine to me.
answered Aug 2 at 14:11
Arthur
98.2k792174
First equivalence is the defnition of $to$, and the second one is De Morgan's law plus double negation. Looks fine to me.
answered Aug 2 at 14:11
Arthur
98.2k792174
answered Aug 2 at 14:11
Arthur
98.2k792174
answered Aug 2 at 14:11
Arthur
98.2k792174
answered Aug 2 at 14:11
Arthur
98.2k792174
98.2k792174
add a comment |Â
add a comment |Â
up vote
0
down vote
You can also observe this easily via the classical method of proof tables in propositional calculus:
beginarray
hline
A& B & neg(Arightarrow B) & neg(neg Alor B) & Alandneg B \ hline
0& 0& 0& 0& 0\ hline
0& 1& 0& 0& 0\ hline
1& 0& 1& 1& 1\ hline
1& 1& 0& 0& 0\ hline
endarray
by using the classical definitions for the elementary operators:
- Negation: beginarray hline A& neg A \ hline 0& 1\ hline 1& 0\
hline endarray - Conjunction: beginarray
hline
A& B & Aland B\ hline
0& 0& 0\ hline
0& 1& 0\ hline
1& 0& 0\ hline
1& 1& 1\ hline
endarray - Disjunction: beginarray
hline
A& B & Alor B\ hline
0& 0& 0\ hline
0& 1& 1\ hline
1& 0& 1\ hline
1& 1& 1\ hline
endarray - Implication: beginarray
hline
A& B & Arightarrow B\ hline
0& 0& 1\ hline
0& 1& 1\ hline
1& 0& 0\ hline
1& 1& 1\ hline
endarray
Thus, as $Cequiv D$ is valid if and only if the have the same truth value under every interpretation, we get the result.
answered Aug 2 at 14:19
zzuussee
1,142419
add a comment |Â
up vote
0
down vote
You can also observe this easily via the classical method of proof tables in propositional calculus:
beginarray
hline
A& B & neg(Arightarrow B) & neg(neg Alor B) & Alandneg B \ hline
0& 0& 0& 0& 0\ hline
0& 1& 0& 0& 0\ hline
1& 0& 1& 1& 1\ hline
1& 1& 0& 0& 0\ hline
endarray
by using the classical definitions for the elementary operators:
- Negation: beginarray hline A& neg A \ hline 0& 1\ hline 1& 0\
hline endarray - Conjunction: beginarray
hline
A& B & Aland B\ hline
0& 0& 0\ hline
0& 1& 0\ hline
1& 0& 0\ hline
1& 1& 1\ hline
endarray - Disjunction: beginarray
hline
A& B & Alor B\ hline
0& 0& 0\ hline
0& 1& 1\ hline
1& 0& 1\ hline
1& 1& 1\ hline
endarray - Implication: beginarray
hline
A& B & Arightarrow B\ hline
0& 0& 1\ hline
0& 1& 1\ hline
1& 0& 0\ hline
1& 1& 1\ hline
endarray
Thus, as $Cequiv D$ is valid if and only if the have the same truth value under every interpretation, we get the result.
answered Aug 2 at 14:19
zzuussee
1,142419
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can also observe this easily via the classical method of proof tables in propositional calculus:
beginarray
hline
A& B & neg(Arightarrow B) & neg(neg Alor B) & Alandneg B \ hline
0& 0& 0& 0& 0\ hline
0& 1& 0& 0& 0\ hline
1& 0& 1& 1& 1\ hline
1& 1& 0& 0& 0\ hline
endarray
by using the classical definitions for the elementary operators:
- Negation: beginarray hline A& neg A \ hline 0& 1\ hline 1& 0\
hline endarray - Conjunction: beginarray
hline
A& B & Aland B\ hline
0& 0& 0\ hline
0& 1& 0\ hline
1& 0& 0\ hline
1& 1& 1\ hline
endarray - Disjunction: beginarray
hline
A& B & Alor B\ hline
0& 0& 0\ hline
0& 1& 1\ hline
1& 0& 1\ hline
1& 1& 1\ hline
endarray - Implication: beginarray
hline
A& B & Arightarrow B\ hline
0& 0& 1\ hline
0& 1& 1\ hline
1& 0& 0\ hline
1& 1& 1\ hline
endarray
Thus, as $Cequiv D$ is valid if and only if the have the same truth value under every interpretation, we get the result.
answered Aug 2 at 14:19
zzuussee
1,142419
You can also observe this easily via the classical method of proof tables in propositional calculus:
beginarray
hline
A& B & neg(Arightarrow B) & neg(neg Alor B) & Alandneg B \ hline
0& 0& 0& 0& 0\ hline
0& 1& 0& 0& 0\ hline
1& 0& 1& 1& 1\ hline
1& 1& 0& 0& 0\ hline
endarray
by using the classical definitions for the elementary operators:
- Negation: beginarray hline A& neg A \ hline 0& 1\ hline 1& 0\
hline endarray - Conjunction: beginarray
hline
A& B & Aland B\ hline
0& 0& 0\ hline
0& 1& 0\ hline
1& 0& 0\ hline
1& 1& 1\ hline
endarray - Disjunction: beginarray
hline
A& B & Alor B\ hline
0& 0& 0\ hline
0& 1& 1\ hline
1& 0& 1\ hline
1& 1& 1\ hline
endarray - Implication: beginarray
hline
A& B & Arightarrow B\ hline
0& 0& 1\ hline
0& 1& 1\ hline
1& 0& 0\ hline
1& 1& 1\ hline
endarray
Thus, as $Cequiv D$ is valid if and only if the have the same truth value under every interpretation, we get the result.
answered Aug 2 at 14:19
zzuussee
1,142419
edited Aug 2 at 14:40
answered Aug 2 at 14:19
zzuussee
1,142419
answered Aug 2 at 14:19
zzuussee
1,142419
answered Aug 2 at 14:19
zzuussee
1,142419
1,142419
add a comment |Â
add a comment |Â
up vote
0
down vote
Yes or no, depending on exactly what you mean.
When people write $Pequiv Qequiv R$ in an informal context they usually mean that $P$ is equivalent to $Q$ and also $Q$ is equivalent to $R$. If that's what you meant then the answer to your question is yes, all three of those formulas are equivalent (if this is not clear to you you really haven't been paying attention in class - it's trivial to verify from truth tables).
But writing $Pequiv Qequiv R$ with the intended meaning as above is really bad notation, especially if, as here, $P$, $Q$ and $R$ are wffs in some formal system. Because then it looks as though $Pequiv Qequiv R$ is meant as a single formula in that system; if so it's not well formed, needing to be parenthesized as $(Pequiv Q)equiv R$ or $Pequiv (Qequiv R)$, neither of which "means" $P$ is equivalent to $Q$ and $Q$ is equivalent to $R$.
So: In the title you refer to "[t]his logical statement". If you mean to be asking about $neg(Arightarrow B)equiv neg(neg Avee B)equiv (Awedge neg B)$ as a wff in propositional logic then it is certainly not valid; neither of the interpretations $(neg(Arightarrow B)equiv neg(neg Avee B))equiv (Awedge neg B)$ and $neg(Arightarrow B)equiv (neg(neg Avee B)equiv (Awedge neg B))$ is valid, since they both say a tautology is equivalent to a non-tautology.
Moral This issue is why people write "The following are equivalent: $P$, $Q$, $R$" instead of writing $Pequiv Qequiv R$.
answered Aug 2 at 15:56
David C. Ullrich
53.8k33481
add a comment |Â
up vote
0
down vote
Yes or no, depending on exactly what you mean.
When people write $Pequiv Qequiv R$ in an informal context they usually mean that $P$ is equivalent to $Q$ and also $Q$ is equivalent to $R$. If that's what you meant then the answer to your question is yes, all three of those formulas are equivalent (if this is not clear to you you really haven't been paying attention in class - it's trivial to verify from truth tables).
But writing $Pequiv Qequiv R$ with the intended meaning as above is really bad notation, especially if, as here, $P$, $Q$ and $R$ are wffs in some formal system. Because then it looks as though $Pequiv Qequiv R$ is meant as a single formula in that system; if so it's not well formed, needing to be parenthesized as $(Pequiv Q)equiv R$ or $Pequiv (Qequiv R)$, neither of which "means" $P$ is equivalent to $Q$ and $Q$ is equivalent to $R$.
So: In the title you refer to "[t]his logical statement". If you mean to be asking about $neg(Arightarrow B)equiv neg(neg Avee B)equiv (Awedge neg B)$ as a wff in propositional logic then it is certainly not valid; neither of the interpretations $(neg(Arightarrow B)equiv neg(neg Avee B))equiv (Awedge neg B)$ and $neg(Arightarrow B)equiv (neg(neg Avee B)equiv (Awedge neg B))$ is valid, since they both say a tautology is equivalent to a non-tautology.
Moral This issue is why people write "The following are equivalent: $P$, $Q$, $R$" instead of writing $Pequiv Qequiv R$.
answered Aug 2 at 15:56
David C. Ullrich
53.8k33481
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes or no, depending on exactly what you mean.
When people write $Pequiv Qequiv R$ in an informal context they usually mean that $P$ is equivalent to $Q$ and also $Q$ is equivalent to $R$. If that's what you meant then the answer to your question is yes, all three of those formulas are equivalent (if this is not clear to you you really haven't been paying attention in class - it's trivial to verify from truth tables).
But writing $Pequiv Qequiv R$ with the intended meaning as above is really bad notation, especially if, as here, $P$, $Q$ and $R$ are wffs in some formal system. Because then it looks as though $Pequiv Qequiv R$ is meant as a single formula in that system; if so it's not well formed, needing to be parenthesized as $(Pequiv Q)equiv R$ or $Pequiv (Qequiv R)$, neither of which "means" $P$ is equivalent to $Q$ and $Q$ is equivalent to $R$.
So: In the title you refer to "[t]his logical statement". If you mean to be asking about $neg(Arightarrow B)equiv neg(neg Avee B)equiv (Awedge neg B)$ as a wff in propositional logic then it is certainly not valid; neither of the interpretations $(neg(Arightarrow B)equiv neg(neg Avee B))equiv (Awedge neg B)$ and $neg(Arightarrow B)equiv (neg(neg Avee B)equiv (Awedge neg B))$ is valid, since they both say a tautology is equivalent to a non-tautology.
Moral This issue is why people write "The following are equivalent: $P$, $Q$, $R$" instead of writing $Pequiv Qequiv R$.
answered Aug 2 at 15:56
David C. Ullrich
53.8k33481
Yes or no, depending on exactly what you mean.
When people write $Pequiv Qequiv R$ in an informal context they usually mean that $P$ is equivalent to $Q$ and also $Q$ is equivalent to $R$. If that's what you meant then the answer to your question is yes, all three of those formulas are equivalent (if this is not clear to you you really haven't been paying attention in class - it's trivial to verify from truth tables).
But writing $Pequiv Qequiv R$ with the intended meaning as above is really bad notation, especially if, as here, $P$, $Q$ and $R$ are wffs in some formal system. Because then it looks as though $Pequiv Qequiv R$ is meant as a single formula in that system; if so it's not well formed, needing to be parenthesized as $(Pequiv Q)equiv R$ or $Pequiv (Qequiv R)$, neither of which "means" $P$ is equivalent to $Q$ and $Q$ is equivalent to $R$.
So: In the title you refer to "[t]his logical statement". If you mean to be asking about $neg(Arightarrow B)equiv neg(neg Avee B)equiv (Awedge neg B)$ as a wff in propositional logic then it is certainly not valid; neither of the interpretations $(neg(Arightarrow B)equiv neg(neg Avee B))equiv (Awedge neg B)$ and $neg(Arightarrow B)equiv (neg(neg Avee B)equiv (Awedge neg B))$ is valid, since they both say a tautology is equivalent to a non-tautology.
Moral This issue is why people write "The following are equivalent: $P$, $Q$, $R$" instead of writing $Pequiv Qequiv R$.
answered Aug 2 at 15:56
David C. Ullrich
53.8k33481
edited Aug 2 at 16:03
answered Aug 2 at 15:56
David C. Ullrich
53.8k33481
answered Aug 2 at 15:56
David C. Ullrich
53.8k33481
answered Aug 2 at 15:56
David C. Ullrich
53.8k33481
53.8k33481
add a comment |Â
add a comment |Â
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1
I just checked this and I could not find any mistake so I guess yes.
– mrtaurho
Aug 2 at 14:08
1
It is not really proof verification though.
– zzuussee
Aug 2 at 14:19