Mixing
Show that the boolean formulas $[(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r] and (p ∧ ¬q) ∨ (r ∧ q)$...
$begingroup$
So far I got this:
$[(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r]$:
- $p∧ T ∧ [(¬q∧p) ∨ r]$
- $p∧ [(p∧¬q) ∨ r]$
- $p lor r$
$(p ∧ ¬q) ∨ (r ∧ q):$
$(p ∧ ¬q) ∨ (q ∧ r)$
$(p ∧( ¬q ∨ q)∧ r$
$p ∧ T ∧ r$
$p ∧ r$
propositional-calculus
edited Jan 24 at 2:19
Namaste
1
asked Jan 24 at 0:03
Derek LongDerek Long
115
$endgroup$
add a comment |
$begingroup$
So far I got this:
$[(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r]$:
- $p∧ T ∧ [(¬q∧p) ∨ r]$
- $p∧ [(p∧¬q) ∨ r]$
- $p lor r$
$(p ∧ ¬q) ∨ (r ∧ q):$
$(p ∧ ¬q) ∨ (q ∧ r)$
$(p ∧( ¬q ∨ q)∧ r$
$p ∧ T ∧ r$
$p ∧ r$
propositional-calculus
edited Jan 24 at 2:19
Namaste
1
asked Jan 24 at 0:03
Derek LongDerek Long
115
$endgroup$
1
$begingroup$
By commutativity of $land$, your first formula is equivalent to $[(plandneg q)lor q]land[(plandneg q)lor r]$ and your second formula is equivalent to $(plandneg q)lor(qland r)$. And these are equivalent by the distributive law.
$endgroup$
– Andreas Blass
Jan 24 at 0:11
$begingroup$
can you please explain a little more on how [(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r] =(p ∧ ¬q) ∨ (r ∧ q)
$endgroup$
– Derek Long
Jan 24 at 1:22
$begingroup$
[(p∧¬q) V q ] ∧(p∧¬q) v r ] under first formula?
$endgroup$
– Derek Long
Jan 24 at 1:30
$begingroup$
Unfortunately, I don't see a way to explain more. My comment already broke the argument into three steps, each applying only one of the laws of propositional logic --- two uses of the commutative law of $land$ and one use of the distributive law. I see no way to break it down any further.
$endgroup$
– Andreas Blass
Jan 24 at 11:30
$begingroup$
Thank you! its making a lot more sense!
$endgroup$
– Derek Long
Jan 24 at 16:29
add a comment |
$begingroup$
So far I got this:
$[(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r]$:
- $p∧ T ∧ [(¬q∧p) ∨ r]$
- $p∧ [(p∧¬q) ∨ r]$
- $p lor r$
$(p ∧ ¬q) ∨ (r ∧ q):$
$(p ∧ ¬q) ∨ (q ∧ r)$
$(p ∧( ¬q ∨ q)∧ r$
$p ∧ T ∧ r$
$p ∧ r$
propositional-calculus
edited Jan 24 at 2:19
Namaste
1
asked Jan 24 at 0:03
Derek LongDerek Long
115
$endgroup$
So far I got this:
$[(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r]$:
- $p∧ T ∧ [(¬q∧p) ∨ r]$
- $p∧ [(p∧¬q) ∨ r]$
- $p lor r$
$(p ∧ ¬q) ∨ (r ∧ q):$
$(p ∧ ¬q) ∨ (q ∧ r)$
$(p ∧( ¬q ∨ q)∧ r$
$p ∧ T ∧ r$
$p ∧ r$
propositional-calculus
propositional-calculus
edited Jan 24 at 2:19
Namaste
1
asked Jan 24 at 0:03
Derek LongDerek Long
115
edited Jan 24 at 2:19
Namaste
1
asked Jan 24 at 0:03
Derek LongDerek Long
115
edited Jan 24 at 2:19
Namaste
1
edited Jan 24 at 2:19
Namaste
1
edited Jan 24 at 2:19
Namaste
1
1
asked Jan 24 at 0:03
Derek LongDerek Long
115
asked Jan 24 at 0:03
Derek LongDerek Long
115
asked Jan 24 at 0:03
Derek LongDerek Long
115
115
1
$begingroup$
By commutativity of $land$, your first formula is equivalent to $[(plandneg q)lor q]land[(plandneg q)lor r]$ and your second formula is equivalent to $(plandneg q)lor(qland r)$. And these are equivalent by the distributive law.
$endgroup$
– Andreas Blass
Jan 24 at 0:11
$begingroup$
can you please explain a little more on how [(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r] =(p ∧ ¬q) ∨ (r ∧ q)
$endgroup$
– Derek Long
Jan 24 at 1:22
$begingroup$
[(p∧¬q) V q ] ∧(p∧¬q) v r ] under first formula?
$endgroup$
– Derek Long
Jan 24 at 1:30
$begingroup$
Unfortunately, I don't see a way to explain more. My comment already broke the argument into three steps, each applying only one of the laws of propositional logic --- two uses of the commutative law of $land$ and one use of the distributive law. I see no way to break it down any further.
$endgroup$
– Andreas Blass
Jan 24 at 11:30
$begingroup$
Thank you! its making a lot more sense!
$endgroup$
– Derek Long
Jan 24 at 16:29
add a comment |
1
$begingroup$
By commutativity of $land$, your first formula is equivalent to $[(plandneg q)lor q]land[(plandneg q)lor r]$ and your second formula is equivalent to $(plandneg q)lor(qland r)$. And these are equivalent by the distributive law.
$endgroup$
– Andreas Blass
Jan 24 at 0:11
$begingroup$
can you please explain a little more on how [(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r] =(p ∧ ¬q) ∨ (r ∧ q)
$endgroup$
– Derek Long
Jan 24 at 1:22
$begingroup$
[(p∧¬q) V q ] ∧(p∧¬q) v r ] under first formula?
$endgroup$
– Derek Long
Jan 24 at 1:30
$begingroup$
Unfortunately, I don't see a way to explain more. My comment already broke the argument into three steps, each applying only one of the laws of propositional logic --- two uses of the commutative law of $land$ and one use of the distributive law. I see no way to break it down any further.
$endgroup$
– Andreas Blass
Jan 24 at 11:30
$begingroup$
Thank you! its making a lot more sense!
$endgroup$
– Derek Long
Jan 24 at 16:29
1
1
$begingroup$
By commutativity of $land$, your first formula is equivalent to $[(plandneg q)lor q]land[(plandneg q)lor r]$ and your second formula is equivalent to $(plandneg q)lor(qland r)$. And these are equivalent by the distributive law.
$endgroup$
– Andreas Blass
Jan 24 at 0:11
$begingroup$
By commutativity of $land$, your first formula is equivalent to $[(plandneg q)lor q]land[(plandneg q)lor r]$ and your second formula is equivalent to $(plandneg q)lor(qland r)$. And these are equivalent by the distributive law.
$endgroup$
– Andreas Blass
Jan 24 at 0:11
$begingroup$
can you please explain a little more on how [(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r] =(p ∧ ¬q) ∨ (r ∧ q)
$endgroup$
– Derek Long
Jan 24 at 1:22
$begingroup$
can you please explain a little more on how [(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r] =(p ∧ ¬q) ∨ (r ∧ q)
$endgroup$
– Derek Long
Jan 24 at 1:22
$begingroup$
[(p∧¬q) V q ] ∧(p∧¬q) v r ] under first formula?
$endgroup$
– Derek Long
Jan 24 at 1:30
$begingroup$
[(p∧¬q) V q ] ∧(p∧¬q) v r ] under first formula?
$endgroup$
– Derek Long
Jan 24 at 1:30
$begingroup$
Unfortunately, I don't see a way to explain more. My comment already broke the argument into three steps, each applying only one of the laws of propositional logic --- two uses of the commutative law of $land$ and one use of the distributive law. I see no way to break it down any further.
$endgroup$
– Andreas Blass
Jan 24 at 11:30
$begingroup$
Unfortunately, I don't see a way to explain more. My comment already broke the argument into three steps, each applying only one of the laws of propositional logic --- two uses of the commutative law of $land$ and one use of the distributive law. I see no way to break it down any further.
$endgroup$
– Andreas Blass
Jan 24 at 11:30
$begingroup$
Thank you! its making a lot more sense!
$endgroup$
– Derek Long
Jan 24 at 16:29
$begingroup$
Thank you! its making a lot more sense!
$endgroup$
– Derek Long
Jan 24 at 16:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're not doing this right. You can't just move parentheses around like you do.
For example, for the first one, you go from $(p land neg q) lor q$ to $p land top$, but that must mean you went from $(p land neg q) lor q$ to $p land (neg q lor q)$ .... which is not right:
In general, $(p land q) lor r$ is not equivalent to $p land (q lor r)$
You are similarly moving parentheses for the second expression in a way that is not allowed. Think about it: what if this was an algebraic expression using numbers? It would be like going from $(3 + 4) cdot (5+6)$ to $3 + 4 cdot 5 + 6$ ... that's clearly not something you can do!
answered Jan 24 at 0:20
Bram28Bram28
63.7k44793
$endgroup$
$begingroup$
yea that makes sense, im just not sure where to begin.
$endgroup$
– Derek Long
Jan 24 at 1:05
$begingroup$
@DerekLong see Andreas' comment !
$endgroup$
– Bram28
Jan 24 at 1:17
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
You're not doing this right. You can't just move parentheses around like you do.
For example, for the first one, you go from $(p land neg q) lor q$ to $p land top$, but that must mean you went from $(p land neg q) lor q$ to $p land (neg q lor q)$ .... which is not right:
In general, $(p land q) lor r$ is not equivalent to $p land (q lor r)$
You are similarly moving parentheses for the second expression in a way that is not allowed. Think about it: what if this was an algebraic expression using numbers? It would be like going from $(3 + 4) cdot (5+6)$ to $3 + 4 cdot 5 + 6$ ... that's clearly not something you can do!
answered Jan 24 at 0:20
Bram28Bram28
63.7k44793
$endgroup$
$begingroup$
yea that makes sense, im just not sure where to begin.
$endgroup$
– Derek Long
Jan 24 at 1:05
$begingroup$
@DerekLong see Andreas' comment !
$endgroup$
– Bram28
Jan 24 at 1:17
add a comment |
$begingroup$
You're not doing this right. You can't just move parentheses around like you do.
For example, for the first one, you go from $(p land neg q) lor q$ to $p land top$, but that must mean you went from $(p land neg q) lor q$ to $p land (neg q lor q)$ .... which is not right:
In general, $(p land q) lor r$ is not equivalent to $p land (q lor r)$
You are similarly moving parentheses for the second expression in a way that is not allowed. Think about it: what if this was an algebraic expression using numbers? It would be like going from $(3 + 4) cdot (5+6)$ to $3 + 4 cdot 5 + 6$ ... that's clearly not something you can do!
answered Jan 24 at 0:20
Bram28Bram28
63.7k44793
$endgroup$
$begingroup$
yea that makes sense, im just not sure where to begin.
$endgroup$
– Derek Long
Jan 24 at 1:05
$begingroup$
@DerekLong see Andreas' comment !
$endgroup$
– Bram28
Jan 24 at 1:17
add a comment |
$begingroup$
You're not doing this right. You can't just move parentheses around like you do.
For example, for the first one, you go from $(p land neg q) lor q$ to $p land top$, but that must mean you went from $(p land neg q) lor q$ to $p land (neg q lor q)$ .... which is not right:
In general, $(p land q) lor r$ is not equivalent to $p land (q lor r)$
You are similarly moving parentheses for the second expression in a way that is not allowed. Think about it: what if this was an algebraic expression using numbers? It would be like going from $(3 + 4) cdot (5+6)$ to $3 + 4 cdot 5 + 6$ ... that's clearly not something you can do!
answered Jan 24 at 0:20
Bram28Bram28
63.7k44793
$endgroup$
You're not doing this right. You can't just move parentheses around like you do.
For example, for the first one, you go from $(p land neg q) lor q$ to $p land top$, but that must mean you went from $(p land neg q) lor q$ to $p land (neg q lor q)$ .... which is not right:
In general, $(p land q) lor r$ is not equivalent to $p land (q lor r)$
You are similarly moving parentheses for the second expression in a way that is not allowed. Think about it: what if this was an algebraic expression using numbers? It would be like going from $(3 + 4) cdot (5+6)$ to $3 + 4 cdot 5 + 6$ ... that's clearly not something you can do!
answered Jan 24 at 0:20
Bram28Bram28
63.7k44793
edited Jan 24 at 0:32
answered Jan 24 at 0:20
Bram28Bram28
63.7k44793
answered Jan 24 at 0:20
Bram28Bram28
63.7k44793
answered Jan 24 at 0:20
Bram28Bram28
63.7k44793
63.7k44793
$begingroup$
yea that makes sense, im just not sure where to begin.
$endgroup$
– Derek Long
Jan 24 at 1:05
$begingroup$
@DerekLong see Andreas' comment !
$endgroup$
– Bram28
Jan 24 at 1:17
add a comment |
$begingroup$
yea that makes sense, im just not sure where to begin.
$endgroup$
– Derek Long
Jan 24 at 1:05
$begingroup$
@DerekLong see Andreas' comment !
$endgroup$
– Bram28
Jan 24 at 1:17
$begingroup$
yea that makes sense, im just not sure where to begin.
$endgroup$
– Derek Long
Jan 24 at 1:05
$begingroup$
yea that makes sense, im just not sure where to begin.
$endgroup$
– Derek Long
Jan 24 at 1:05
$begingroup$
@DerekLong see Andreas' comment !
$endgroup$
– Bram28
Jan 24 at 1:17
$begingroup$
@DerekLong see Andreas' comment !
$endgroup$
– Bram28
Jan 24 at 1:17
add a comment |
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1
$begingroup$
By commutativity of $land$, your first formula is equivalent to $[(plandneg q)lor q]land[(plandneg q)lor r]$ and your second formula is equivalent to $(plandneg q)lor(qland r)$. And these are equivalent by the distributive law.
$endgroup$
– Andreas Blass
Jan 24 at 0:11
$begingroup$
can you please explain a little more on how [(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r] =(p ∧ ¬q) ∨ (r ∧ q)
$endgroup$
– Derek Long
Jan 24 at 1:22
$begingroup$
[(p∧¬q) V q ] ∧(p∧¬q) v r ] under first formula?
$endgroup$
– Derek Long
Jan 24 at 1:30
$begingroup$
Unfortunately, I don't see a way to explain more. My comment already broke the argument into three steps, each applying only one of the laws of propositional logic --- two uses of the commutative law of $land$ and one use of the distributive law. I see no way to break it down any further.
$endgroup$
– Andreas Blass
Jan 24 at 11:30
$begingroup$
Thank you! its making a lot more sense!
$endgroup$
– Derek Long
Jan 24 at 16:29