Mixing
Proving Expression is Contingency with Logical Equivalences
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I am trying to prove that $lnot[(q vee r) wedge ((p vee q) wedge (lnot p vee r))]$ is a contingency using logical equivalence rules.
I have tried various steps and keep getting stuck in loops, or making wrong moves and getting tautology or contradiction.
The last steps I tried were dropping the brackets surrounding all the or's (commutativity) and getting $lnot[q vee r vee p vee q wedge (lnot pvee r)]$ then followed by:
$$ lnot[q vee p vee r wedge (lnot pvee r)] mbox{ (idempotent)} $$
$$lnot[q vee p vee r wedge (lnot p vee r)] mbox{ (commutativity again)} $$
$$lnot(q vee p vee r) mbox{ (absorption)}
$$
For one I feel like I'm making some either useless, or incorrect steps here, as well I don't see how $lnot(q vee p vee r)$ would prove it's a contingency. I've tried looking at examples and other questions related to this topic, but I just can't seem to find anything that makes sense given the question. Any help would be greatly appreciated, thank you!
logic propositional-calculus
asked Jan 27 at 18:28
NickNick
85
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add a comment |
$begingroup$
I am trying to prove that $lnot[(q vee r) wedge ((p vee q) wedge (lnot p vee r))]$ is a contingency using logical equivalence rules.
I have tried various steps and keep getting stuck in loops, or making wrong moves and getting tautology or contradiction.
The last steps I tried were dropping the brackets surrounding all the or's (commutativity) and getting $lnot[q vee r vee p vee q wedge (lnot pvee r)]$ then followed by:
$$ lnot[q vee p vee r wedge (lnot pvee r)] mbox{ (idempotent)} $$
$$lnot[q vee p vee r wedge (lnot p vee r)] mbox{ (commutativity again)} $$
$$lnot(q vee p vee r) mbox{ (absorption)}
$$
For one I feel like I'm making some either useless, or incorrect steps here, as well I don't see how $lnot(q vee p vee r)$ would prove it's a contingency. I've tried looking at examples and other questions related to this topic, but I just can't seem to find anything that makes sense given the question. Any help would be greatly appreciated, thank you!
logic propositional-calculus
asked Jan 27 at 18:28
NickNick
85
$endgroup$
add a comment |
$begingroup$
I am trying to prove that $lnot[(q vee r) wedge ((p vee q) wedge (lnot p vee r))]$ is a contingency using logical equivalence rules.
I have tried various steps and keep getting stuck in loops, or making wrong moves and getting tautology or contradiction.
The last steps I tried were dropping the brackets surrounding all the or's (commutativity) and getting $lnot[q vee r vee p vee q wedge (lnot pvee r)]$ then followed by:
$$ lnot[q vee p vee r wedge (lnot pvee r)] mbox{ (idempotent)} $$
$$lnot[q vee p vee r wedge (lnot p vee r)] mbox{ (commutativity again)} $$
$$lnot(q vee p vee r) mbox{ (absorption)}
$$
For one I feel like I'm making some either useless, or incorrect steps here, as well I don't see how $lnot(q vee p vee r)$ would prove it's a contingency. I've tried looking at examples and other questions related to this topic, but I just can't seem to find anything that makes sense given the question. Any help would be greatly appreciated, thank you!
logic propositional-calculus
asked Jan 27 at 18:28
NickNick
85
$endgroup$
I am trying to prove that $lnot[(q vee r) wedge ((p vee q) wedge (lnot p vee r))]$ is a contingency using logical equivalence rules.
I have tried various steps and keep getting stuck in loops, or making wrong moves and getting tautology or contradiction.
The last steps I tried were dropping the brackets surrounding all the or's (commutativity) and getting $lnot[q vee r vee p vee q wedge (lnot pvee r)]$ then followed by:
$$ lnot[q vee p vee r wedge (lnot pvee r)] mbox{ (idempotent)} $$
$$lnot[q vee p vee r wedge (lnot p vee r)] mbox{ (commutativity again)} $$
$$lnot(q vee p vee r) mbox{ (absorption)}
$$
For one I feel like I'm making some either useless, or incorrect steps here, as well I don't see how $lnot(q vee p vee r)$ would prove it's a contingency. I've tried looking at examples and other questions related to this topic, but I just can't seem to find anything that makes sense given the question. Any help would be greatly appreciated, thank you!
logic propositional-calculus
logic propositional-calculus
asked Jan 27 at 18:28
NickNick
85
asked Jan 27 at 18:28
NickNick
85
edited Jan 27 at 20:27
Nick
asked Jan 27 at 18:28
NickNick
85
asked Jan 27 at 18:28
NickNick
85
asked Jan 27 at 18:28
NickNick
85
85
add a comment |
add a comment |
1 Answer
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To prove a statement is a contingency you need to show that it is possible for the statement to be true, as well as show that it is possible for the statement to be false.
Now, it doesn't take much experience with logic to recognize $neg (q lor p lor r)$ as a contingency, so for some audiences what you did will be enough. However, for a hard proof, you probably want to come up with two different truth-valuie assignment: one that sets the statement to True, and another on that sets the statement to false.
For example, to set the statement to false, we can set $p=q=r=True$, for then the statement evaluates to $neg (q lor p lor r) =neg (T lor T lor T) = neg T = F$
Can you come up with a truth-value assignment that makes the statement True?
answered Jan 27 at 19:03
Bram28Bram28
63.9k44793
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$begingroup$
Okay, I wasn't sure about using assignments to prove its state, so thank you. :) And values that would make the statement True would be if all of them were False, right?
$endgroup$
– Nick
Jan 27 at 19:10
$begingroup$
@Nick Correct! That's all you need to do ... interestingly, you could have done that with the original statement, i.e. without doing any algebra at all ... but it would have been harder to find those two assignments. With your simplified expression, it's a piece of cake! :)
$endgroup$
– Bram28
Jan 27 at 19:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
To prove a statement is a contingency you need to show that it is possible for the statement to be true, as well as show that it is possible for the statement to be false.
Now, it doesn't take much experience with logic to recognize $neg (q lor p lor r)$ as a contingency, so for some audiences what you did will be enough. However, for a hard proof, you probably want to come up with two different truth-valuie assignment: one that sets the statement to True, and another on that sets the statement to false.
For example, to set the statement to false, we can set $p=q=r=True$, for then the statement evaluates to $neg (q lor p lor r) =neg (T lor T lor T) = neg T = F$
Can you come up with a truth-value assignment that makes the statement True?
answered Jan 27 at 19:03
Bram28Bram28
63.9k44793
$endgroup$
$begingroup$
Okay, I wasn't sure about using assignments to prove its state, so thank you. :) And values that would make the statement True would be if all of them were False, right?
$endgroup$
– Nick
Jan 27 at 19:10
$begingroup$
@Nick Correct! That's all you need to do ... interestingly, you could have done that with the original statement, i.e. without doing any algebra at all ... but it would have been harder to find those two assignments. With your simplified expression, it's a piece of cake! :)
$endgroup$
– Bram28
Jan 27 at 19:12
add a comment |
$begingroup$
To prove a statement is a contingency you need to show that it is possible for the statement to be true, as well as show that it is possible for the statement to be false.
Now, it doesn't take much experience with logic to recognize $neg (q lor p lor r)$ as a contingency, so for some audiences what you did will be enough. However, for a hard proof, you probably want to come up with two different truth-valuie assignment: one that sets the statement to True, and another on that sets the statement to false.
For example, to set the statement to false, we can set $p=q=r=True$, for then the statement evaluates to $neg (q lor p lor r) =neg (T lor T lor T) = neg T = F$
Can you come up with a truth-value assignment that makes the statement True?
answered Jan 27 at 19:03
Bram28Bram28
63.9k44793
$endgroup$
$begingroup$
Okay, I wasn't sure about using assignments to prove its state, so thank you. :) And values that would make the statement True would be if all of them were False, right?
$endgroup$
– Nick
Jan 27 at 19:10
$begingroup$
@Nick Correct! That's all you need to do ... interestingly, you could have done that with the original statement, i.e. without doing any algebra at all ... but it would have been harder to find those two assignments. With your simplified expression, it's a piece of cake! :)
$endgroup$
– Bram28
Jan 27 at 19:12
add a comment |
$begingroup$
To prove a statement is a contingency you need to show that it is possible for the statement to be true, as well as show that it is possible for the statement to be false.
Now, it doesn't take much experience with logic to recognize $neg (q lor p lor r)$ as a contingency, so for some audiences what you did will be enough. However, for a hard proof, you probably want to come up with two different truth-valuie assignment: one that sets the statement to True, and another on that sets the statement to false.
For example, to set the statement to false, we can set $p=q=r=True$, for then the statement evaluates to $neg (q lor p lor r) =neg (T lor T lor T) = neg T = F$
Can you come up with a truth-value assignment that makes the statement True?
answered Jan 27 at 19:03
Bram28Bram28
63.9k44793
$endgroup$
To prove a statement is a contingency you need to show that it is possible for the statement to be true, as well as show that it is possible for the statement to be false.
Now, it doesn't take much experience with logic to recognize $neg (q lor p lor r)$ as a contingency, so for some audiences what you did will be enough. However, for a hard proof, you probably want to come up with two different truth-valuie assignment: one that sets the statement to True, and another on that sets the statement to false.
For example, to set the statement to false, we can set $p=q=r=True$, for then the statement evaluates to $neg (q lor p lor r) =neg (T lor T lor T) = neg T = F$
Can you come up with a truth-value assignment that makes the statement True?
answered Jan 27 at 19:03
Bram28Bram28
63.9k44793
answered Jan 27 at 19:03
Bram28Bram28
63.9k44793
answered Jan 27 at 19:03
Bram28Bram28
63.9k44793
answered Jan 27 at 19:03
Bram28Bram28
63.9k44793
63.9k44793
$begingroup$
Okay, I wasn't sure about using assignments to prove its state, so thank you. :) And values that would make the statement True would be if all of them were False, right?
$endgroup$
– Nick
Jan 27 at 19:10
$begingroup$
@Nick Correct! That's all you need to do ... interestingly, you could have done that with the original statement, i.e. without doing any algebra at all ... but it would have been harder to find those two assignments. With your simplified expression, it's a piece of cake! :)
$endgroup$
– Bram28
Jan 27 at 19:12
add a comment |
$begingroup$
Okay, I wasn't sure about using assignments to prove its state, so thank you. :) And values that would make the statement True would be if all of them were False, right?
$endgroup$
– Nick
Jan 27 at 19:10
$begingroup$
@Nick Correct! That's all you need to do ... interestingly, you could have done that with the original statement, i.e. without doing any algebra at all ... but it would have been harder to find those two assignments. With your simplified expression, it's a piece of cake! :)
$endgroup$
– Bram28
Jan 27 at 19:12
$begingroup$
Okay, I wasn't sure about using assignments to prove its state, so thank you. :) And values that would make the statement True would be if all of them were False, right?
$endgroup$
– Nick
Jan 27 at 19:10
$begingroup$
Okay, I wasn't sure about using assignments to prove its state, so thank you. :) And values that would make the statement True would be if all of them were False, right?
$endgroup$
– Nick
Jan 27 at 19:10
$begingroup$
@Nick Correct! That's all you need to do ... interestingly, you could have done that with the original statement, i.e. without doing any algebra at all ... but it would have been harder to find those two assignments. With your simplified expression, it's a piece of cake! :)
$endgroup$
– Bram28
Jan 27 at 19:12
$begingroup$
@Nick Correct! That's all you need to do ... interestingly, you could have done that with the original statement, i.e. without doing any algebra at all ... but it would have been harder to find those two assignments. With your simplified expression, it's a piece of cake! :)
$endgroup$
– Bram28
Jan 27 at 19:12
add a comment |
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