Mixing
Propositional Logic Help: $(neg p wedge (p vee q)) rightarrow q $ is a tautology
$begingroup$
I need to prove that $(neg p wedge (p vee q)) rightarrow q $ is a tautology using Laws of Logic (not truth tables).
This is what I tried:
$equiv (( neg p wedge p) vee (neg p wedge q)) rightarrow q \
equiv (F vee (neg p wedge q)) rightarrow q \
equiv (neg p wedge q) rightarrow q \
equiv (F) rightarrow q \
equiv T $
Is this logically correct?
The laws I used in order were: Distributive, then negation, and identity. My only issue is with the last step where I know the truth values of $(neg p wedge q)$ are all $F$ but I dont know what law it uses.
Please Help!
logic propositional-calculus formal-proofs
edited Jan 27 at 5:36
Martin Sleziak
44.9k10122276
asked Jan 31 '15 at 2:14
Digital ShrapnelDigital Shrapnel
556
$endgroup$
|
show 11 more comments
$begingroup$
I need to prove that $(neg p wedge (p vee q)) rightarrow q $ is a tautology using Laws of Logic (not truth tables).
This is what I tried:
$equiv (( neg p wedge p) vee (neg p wedge q)) rightarrow q \
equiv (F vee (neg p wedge q)) rightarrow q \
equiv (neg p wedge q) rightarrow q \
equiv (F) rightarrow q \
equiv T $
Is this logically correct?
The laws I used in order were: Distributive, then negation, and identity. My only issue is with the last step where I know the truth values of $(neg p wedge q)$ are all $F$ but I dont know what law it uses.
Please Help!
logic propositional-calculus formal-proofs
edited Jan 27 at 5:36
Martin Sleziak
44.9k10122276
asked Jan 31 '15 at 2:14
Digital ShrapnelDigital Shrapnel
556
$endgroup$
$begingroup$
$neg pland q$ can have either truth value; it need not be $F$. Do you have $pto qequiv neg plor q$?
$endgroup$
– Brian M. Scott
Jan 31 '15 at 3:08
$begingroup$
As Brian M. Scott points out, $neg pland q$ may be either $T$ or $F$. My answer gives you a strong hint on how to use the equivalences you deduced so far to end up with your tautology. :)
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 3:10
$begingroup$
@induktio Thanks a lot! I forgot about manipulating the $rightarrow$ symbol.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:27
$begingroup$
@DigitalShrapnel Does it make sense now?
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 6:28
$begingroup$
@induktio Yeah. I just need to backtrack to fully understand how $neg p wedge q$ is operated on. But I see where it leads.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:31
|
show 11 more comments
$begingroup$
I need to prove that $(neg p wedge (p vee q)) rightarrow q $ is a tautology using Laws of Logic (not truth tables).
This is what I tried:
$equiv (( neg p wedge p) vee (neg p wedge q)) rightarrow q \
equiv (F vee (neg p wedge q)) rightarrow q \
equiv (neg p wedge q) rightarrow q \
equiv (F) rightarrow q \
equiv T $
Is this logically correct?
The laws I used in order were: Distributive, then negation, and identity. My only issue is with the last step where I know the truth values of $(neg p wedge q)$ are all $F$ but I dont know what law it uses.
Please Help!
logic propositional-calculus formal-proofs
edited Jan 27 at 5:36
Martin Sleziak
44.9k10122276
asked Jan 31 '15 at 2:14
Digital ShrapnelDigital Shrapnel
556
$endgroup$
I need to prove that $(neg p wedge (p vee q)) rightarrow q $ is a tautology using Laws of Logic (not truth tables).
This is what I tried:
$equiv (( neg p wedge p) vee (neg p wedge q)) rightarrow q \
equiv (F vee (neg p wedge q)) rightarrow q \
equiv (neg p wedge q) rightarrow q \
equiv (F) rightarrow q \
equiv T $
Is this logically correct?
The laws I used in order were: Distributive, then negation, and identity. My only issue is with the last step where I know the truth values of $(neg p wedge q)$ are all $F$ but I dont know what law it uses.
Please Help!
logic propositional-calculus formal-proofs
logic propositional-calculus formal-proofs
edited Jan 27 at 5:36
Martin Sleziak
44.9k10122276
asked Jan 31 '15 at 2:14
Digital ShrapnelDigital Shrapnel
556
edited Jan 27 at 5:36
Martin Sleziak
44.9k10122276
asked Jan 31 '15 at 2:14
Digital ShrapnelDigital Shrapnel
556
edited Jan 27 at 5:36
Martin Sleziak
44.9k10122276
edited Jan 27 at 5:36
Martin Sleziak
44.9k10122276
edited Jan 27 at 5:36
Martin Sleziak
44.9k10122276
44.9k10122276
asked Jan 31 '15 at 2:14
Digital ShrapnelDigital Shrapnel
556
asked Jan 31 '15 at 2:14
Digital ShrapnelDigital Shrapnel
556
asked Jan 31 '15 at 2:14
Digital ShrapnelDigital Shrapnel
556
556
$begingroup$
$neg pland q$ can have either truth value; it need not be $F$. Do you have $pto qequiv neg plor q$?
$endgroup$
– Brian M. Scott
Jan 31 '15 at 3:08
$begingroup$
As Brian M. Scott points out, $neg pland q$ may be either $T$ or $F$. My answer gives you a strong hint on how to use the equivalences you deduced so far to end up with your tautology. :)
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 3:10
$begingroup$
@induktio Thanks a lot! I forgot about manipulating the $rightarrow$ symbol.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:27
$begingroup$
@DigitalShrapnel Does it make sense now?
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 6:28
$begingroup$
@induktio Yeah. I just need to backtrack to fully understand how $neg p wedge q$ is operated on. But I see where it leads.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:31
|
show 11 more comments
$begingroup$
$neg pland q$ can have either truth value; it need not be $F$. Do you have $pto qequiv neg plor q$?
$endgroup$
– Brian M. Scott
Jan 31 '15 at 3:08
$begingroup$
As Brian M. Scott points out, $neg pland q$ may be either $T$ or $F$. My answer gives you a strong hint on how to use the equivalences you deduced so far to end up with your tautology. :)
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 3:10
$begingroup$
@induktio Thanks a lot! I forgot about manipulating the $rightarrow$ symbol.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:27
$begingroup$
@DigitalShrapnel Does it make sense now?
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 6:28
$begingroup$
@induktio Yeah. I just need to backtrack to fully understand how $neg p wedge q$ is operated on. But I see where it leads.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:31
$begingroup$
$neg pland q$ can have either truth value; it need not be $F$. Do you have $pto qequiv neg plor q$?
$endgroup$
– Brian M. Scott
Jan 31 '15 at 3:08
$begingroup$
$neg pland q$ can have either truth value; it need not be $F$. Do you have $pto qequiv neg plor q$?
$endgroup$
– Brian M. Scott
Jan 31 '15 at 3:08
$begingroup$
As Brian M. Scott points out, $neg pland q$ may be either $T$ or $F$. My answer gives you a strong hint on how to use the equivalences you deduced so far to end up with your tautology. :)
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 3:10
$begingroup$
As Brian M. Scott points out, $neg pland q$ may be either $T$ or $F$. My answer gives you a strong hint on how to use the equivalences you deduced so far to end up with your tautology. :)
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 3:10
$begingroup$
@induktio Thanks a lot! I forgot about manipulating the $rightarrow$ symbol.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:27
$begingroup$
@induktio Thanks a lot! I forgot about manipulating the $rightarrow$ symbol.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:27
$begingroup$
@DigitalShrapnel Does it make sense now?
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 6:28
$begingroup$
@DigitalShrapnel Does it make sense now?
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 6:28
$begingroup$
@induktio Yeah. I just need to backtrack to fully understand how $neg p wedge q$ is operated on. But I see where it leads.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:31
$begingroup$
@induktio Yeah. I just need to backtrack to fully understand how $neg p wedge q$ is operated on. But I see where it leads.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:31
|
show 11 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Hint: We have that
$$
(neg p land q)to q equiv (plor neg q)lor q equiv p lor (neg qlor q).
$$
You should be able to take it from there.
answered Jan 31 '15 at 3:07
Daniel W. FarlowDaniel W. Farlow
17.7k114588
$endgroup$
add a comment |
$begingroup$
It's been a while since my discrete class, but here's my try. It looks like all you really need to use is the simple Elimination Rule, which states:
$p wedge q rightarrow p$
OR (stay with me)
$p wedge q rightarrow q$
So, to finish off the last part of your problem:
$equiv (neg p wedge q) rightarrow q \
equiv q rightarrow q \$
Or, heck:
$equiv (neg p wedge q) rightarrow q \
equiv neg p rightarrow q \$
Here's a clean, handy cheat sheet that may help you out. One thing I remember is it's easy getting caught up in what the values of p and q may be, when what's important in "proofs" like these is establishing a route to truth. Thinking that way helped me out.
answered Jan 31 '15 at 3:08
ChucklesChuckles
4518
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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$begingroup$
Hint: We have that
$$
(neg p land q)to q equiv (plor neg q)lor q equiv p lor (neg qlor q).
$$
You should be able to take it from there.
answered Jan 31 '15 at 3:07
Daniel W. FarlowDaniel W. Farlow
17.7k114588
$endgroup$
add a comment |
$begingroup$
Hint: We have that
$$
(neg p land q)to q equiv (plor neg q)lor q equiv p lor (neg qlor q).
$$
You should be able to take it from there.
answered Jan 31 '15 at 3:07
Daniel W. FarlowDaniel W. Farlow
17.7k114588
$endgroup$
add a comment |
$begingroup$
Hint: We have that
$$
(neg p land q)to q equiv (plor neg q)lor q equiv p lor (neg qlor q).
$$
You should be able to take it from there.
answered Jan 31 '15 at 3:07
Daniel W. FarlowDaniel W. Farlow
17.7k114588
$endgroup$
Hint: We have that
$$
(neg p land q)to q equiv (plor neg q)lor q equiv p lor (neg qlor q).
$$
You should be able to take it from there.
answered Jan 31 '15 at 3:07
Daniel W. FarlowDaniel W. Farlow
17.7k114588
answered Jan 31 '15 at 3:07
Daniel W. FarlowDaniel W. Farlow
17.7k114588
answered Jan 31 '15 at 3:07
Daniel W. FarlowDaniel W. Farlow
17.7k114588
answered Jan 31 '15 at 3:07
Daniel W. FarlowDaniel W. Farlow
17.7k114588
17.7k114588
add a comment |
add a comment |
$begingroup$
It's been a while since my discrete class, but here's my try. It looks like all you really need to use is the simple Elimination Rule, which states:
$p wedge q rightarrow p$
OR (stay with me)
$p wedge q rightarrow q$
So, to finish off the last part of your problem:
$equiv (neg p wedge q) rightarrow q \
equiv q rightarrow q \$
Or, heck:
$equiv (neg p wedge q) rightarrow q \
equiv neg p rightarrow q \$
Here's a clean, handy cheat sheet that may help you out. One thing I remember is it's easy getting caught up in what the values of p and q may be, when what's important in "proofs" like these is establishing a route to truth. Thinking that way helped me out.
answered Jan 31 '15 at 3:08
ChucklesChuckles
4518
$endgroup$
add a comment |
$begingroup$
It's been a while since my discrete class, but here's my try. It looks like all you really need to use is the simple Elimination Rule, which states:
$p wedge q rightarrow p$
OR (stay with me)
$p wedge q rightarrow q$
So, to finish off the last part of your problem:
$equiv (neg p wedge q) rightarrow q \
equiv q rightarrow q \$
Or, heck:
$equiv (neg p wedge q) rightarrow q \
equiv neg p rightarrow q \$
Here's a clean, handy cheat sheet that may help you out. One thing I remember is it's easy getting caught up in what the values of p and q may be, when what's important in "proofs" like these is establishing a route to truth. Thinking that way helped me out.
answered Jan 31 '15 at 3:08
ChucklesChuckles
4518
$endgroup$
add a comment |
$begingroup$
It's been a while since my discrete class, but here's my try. It looks like all you really need to use is the simple Elimination Rule, which states:
$p wedge q rightarrow p$
OR (stay with me)
$p wedge q rightarrow q$
So, to finish off the last part of your problem:
$equiv (neg p wedge q) rightarrow q \
equiv q rightarrow q \$
Or, heck:
$equiv (neg p wedge q) rightarrow q \
equiv neg p rightarrow q \$
Here's a clean, handy cheat sheet that may help you out. One thing I remember is it's easy getting caught up in what the values of p and q may be, when what's important in "proofs" like these is establishing a route to truth. Thinking that way helped me out.
answered Jan 31 '15 at 3:08
ChucklesChuckles
4518
$endgroup$
It's been a while since my discrete class, but here's my try. It looks like all you really need to use is the simple Elimination Rule, which states:
$p wedge q rightarrow p$
OR (stay with me)
$p wedge q rightarrow q$
So, to finish off the last part of your problem:
$equiv (neg p wedge q) rightarrow q \
equiv q rightarrow q \$
Or, heck:
$equiv (neg p wedge q) rightarrow q \
equiv neg p rightarrow q \$
Here's a clean, handy cheat sheet that may help you out. One thing I remember is it's easy getting caught up in what the values of p and q may be, when what's important in "proofs" like these is establishing a route to truth. Thinking that way helped me out.
answered Jan 31 '15 at 3:08
ChucklesChuckles
4518
edited Jan 31 '15 at 3:26
answered Jan 31 '15 at 3:08
ChucklesChuckles
4518
answered Jan 31 '15 at 3:08
ChucklesChuckles
4518
answered Jan 31 '15 at 3:08
ChucklesChuckles
4518
4518
add a comment |
add a comment |
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$begingroup$
$neg pland q$ can have either truth value; it need not be $F$. Do you have $pto qequiv neg plor q$?
$endgroup$
– Brian M. Scott
Jan 31 '15 at 3:08
$begingroup$
As Brian M. Scott points out, $neg pland q$ may be either $T$ or $F$. My answer gives you a strong hint on how to use the equivalences you deduced so far to end up with your tautology. :)
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 3:10
$begingroup$
@induktio Thanks a lot! I forgot about manipulating the $rightarrow$ symbol.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:27
$begingroup$
@DigitalShrapnel Does it make sense now?
$endgroup$
– Daniel W. Farlow
Jan 31 '15 at 6:28
$begingroup$
@induktio Yeah. I just need to backtrack to fully understand how $neg p wedge q$ is operated on. But I see where it leads.
$endgroup$
– Digital Shrapnel
Jan 31 '15 at 6:31